轻松模拟习题集

B十1八.
榔头剪刀布 (20)

PAT/简单模拟练习集(2),pat模拟习题集

B101八. 榔头剪刀布 (20)

Discription:

世家应该都会玩“锤子剪刀布”的二十二日游:多人还要提交手势,胜负规则如图所示:

澳门葡京备用网址 1

现给出多个人的竞赛记录,请计算双方的胜、平、负次数,并且付诸双方各自出什么样手势的胜算最大。

Input:

输入第二行提交正整数N(<=十5),即双方交锋的次数。随后N行,每行给出一遍交锋的新闻,即甲、乙双方同时提交的的手势。C代表“锤子”、J代表“剪刀”、B代表“布”,第二个字母代表甲方,第3个象征乙方,中间有二个空格。

Output:

出口第贰、2行分别给出甲、乙的胜、平、负次数,数字间以3个空格分隔。第一行提交七个假名,分别代表甲、乙胜球次数最多的手势,中间有3个空格。假设解不唯一,则输出按字母序最小的解。

Sample Input:

10
C J
J B
C B
B B
B C
C C
C B
J B
B C
J J

Sample Output:

5 3 2
2 3 5
B B

 1 #include <cstdio>
 2 
 3 #define MaxSize 3
 4 
 5 int ListA[MaxSize], ListB[MaxSize], Time[MaxSize];
 6 
 7 int main()
 8 {
 9     //freopen("E:\\Temp\\input.txt", "r", stdin);
10 
11     int N;
12     char c1, c2;
13     scanf("%d", &N);
14     for(int i=0; i<N; i++) {
15         getchar();
16         scanf("%c %c", &c1, &c2);
17             if(c1 == 'B') {
18                 if(c2 == 'B') {
19                     Time[1]++;
20                 } else if(c2 == 'C') {
21                     Time[0]++;
22                     ListA[0]++;
23                 } else {
24                     Time[2]++;
25                     ListB[2]++;
26                 }
27             } else if(c1 == 'C') {
28                 if(c2 == 'B') {
29                     Time[2]++;
30                     ListB[0]++;
31                 } else if(c2 == 'C') {
32                     Time[1]++;
33                 } else {
34                     Time[0]++;
35                     ListA[1]++;
36                 }
37             } else {
38                 if(c2 == 'B') {
39                     Time[0]++;
40                     ListA[2]++;
41                 } else if(c2 == 'C') {
42                     Time[2]++;
43                     ListB[1]++;
44                 } else {
45                     Time[1]++;
46                 }
47             }
48     }
49 
50     printf("%d %d %d\n%d %d %d\n", Time[0], Time[1], Time[2], Time[2], Time[1], Time[0]);
51     if(ListA[0] >= ListA[1]) {
52         if(ListA[0] >= ListA[2])
53             printf("B ");
54         else
55             printf("J ");
56     } else {
57         if(ListA[1] >= ListA[2])
58             printf("C ");
59         else
60             printf("J ");
61     }
62     if(ListB[0] >= ListB[1]) {
63         if(ListB[0] >= ListB[2])
64             printf("B\n");
65         else
66             printf("J\n");
67     } else {
68         if(ListB[1] >= ListB[2])
69             printf("C\n");
70         else
71             printf("J\n");
72     }
73 
74     return 0;
75 }

 1 #include <cstdio>
 2 
 3 int change(char c)
 4 {
 5     if(c == 'B')
 6         return 0;
 7     else if(c == 'C')
 8         return 1;
 9     else
10         return 2;
11 }
12 
13 int main()
14 {
15     //freopen("E:\\Temp\\input.txt", "r", stdin);
16 
17     char mp[3] = {'B', 'C', 'J'};
18     int n;
19     scanf("%d", &n);
20     int times_A[3] = {0}, times_B[3] = {0};
21     int hand_A[3] = {0}, hand_B[3] = {0};
22     char c1, c2;
23     int k1, k2;
24     for(int i=0; i<n; i++) {
25         getchar();
26         scanf("%c %c", &c1, &c2);
27         k1 = change(c1);
28         k2 = change(c2);
29         if((k1+1)%3 == k2) {
30             times_A[0]++;
31             times_B[2]++;
32             hand_A[k1]++;
33         } else if(k1 == k2) {
34             times_A[1]++;
35             times_B[1]++;
36         } else {
37             times_A[2]++;
38             times_B[0]++;
39             hand_B[k2]++;
40         }
41     }
42 
43     printf("%d %d %d\n", times_A[0], times_A[1], times_A[2]);
44     printf("%d %d %d\n", times_B[0], times_B[1], times_B[2]);
45     int id1 = 0, id2 = 0;
46     for(int i=0; i<3; i++) {
47         if(hand_A[i] > hand_A[id1])
48             id1 = i;
49         if(hand_B[i] > hand_B[id2])
50             id2 = i;
51     }
52     printf("%c %c\n", mp[id1], mp[id2]);
53 
54     return 0;
55 }

 

A1042. Shuffling Machine (20)

Description:

Shuffling is a procedure used to randomize a deck of playing cards.
Because standard shuffling techniques are seen as weak, and in order to
avoid “inside jobs” where employees collaborate with gamblers by
performing inadequate shuffles, many casinos employ automatic shuffling
machines
. Your task is to simulate a shuffling machine.

The machine shuffles a deck of 54 cards according to a given random
order and repeats for a given number of times. It is assumed that the
initial status of a card deck is in the following order:

S1, S2, …, S13, H1, H2, …, H13, C1, C2, …, C13, D1, D2, …, D13,
J1, J2

where “S” stands for “Spade”, “H” for “Heart”, “C” for “Club”, “D” for
“Diamond”, and “J” for “Joker”. A given order is a permutation of
distinct integers in [1, 54]. If the number at the i-th position is j,
it means to move the card from position i to position j. For example,
suppose we only have 5 cards: S3, H5, C1, D13 and J2. Given a shuffling
order {4, 2, 5, 3, 1}, the result will be: J2, H5, D13, S3, C1. If we
are to repeat the shuffling again, the result will be: C1, H5, S3, J2,
D13.

Input:

Each input file contains one test case. For each case, the first line
contains a positive integer K (<= 20) which is the number of repeat
times. Then the next line contains the given order. All the numbers in a
line are separated by a space.

Output:

For each test case, print the shuffling results in one line. All the
cards are separated by a space, and there must be no extra space at the
end of the line.

Sample Input:

2
36 52 37 38 3 39 40 53 54 41 11 12 13 42 43 44 2 4 23 24 25 26 27 6 7 8
48 49 50 51 9 10 14 15 16 5 17 18 19 1 20 21 22 28 29 30 31 32 33 34 35
45 46 47

Sample Output:

S7 C11 C10 C12 S1 H7 H8 H9 D8 D9 S11 S12 S13 D10 D11 D12 S3 S4 S6 S10 H1
H2 C13 D2 D3 D4 H6 H3 D13 J1 J2 C1 C2 C3 C4 D1 S5 H5 H11 H12 C6 C7 C8 C9
S2 S8 S9 H10 D5 D6 D7 H4 H13 C5

 1 #include <cstdio>
 2 
 3 #define MaxSize 55
 4 
 5 int List1[MaxSize], List2[MaxSize], List3[MaxSize];
 6 
 7 int main()
 8 {
 9     //freopen("E:\\Temp\\input.txt", "r", stdin);
10 
11     int N, temp;
12     scanf("%d", &N);
13     for(int i=1; i<MaxSize; i++) {
14         List1[i] = i;
15         scanf("%d", &List2[i]);
16     }
17     for(int i=0; i<N; i++) {
18         for(int j=1; j<MaxSize; j++) {
19             List3[List2[j]] = List1[j];
20         }
21         for(int k=1; k<MaxSize; k++) {
22             List1[k] = List3[k];
23         }
24     }
25 
26     for(int i=1; i<MaxSize-1; i++) {
27         if(List1[i] > 52) {
28             printf("J%d ", List1[i]-52);
29         } else if(List1[i] > 39) {
30             printf("D%d ", List1[i]-39);
31         } else if(List1[i] > 26) {
32             printf("C%d ", List1[i]-26);
33         } else if(List1[i] > 13) {
34             printf("H%d ", List1[i]-13);
35         } else {
36             printf("S%d ", List1[i]);
37         }
38     }
39     if(List1[MaxSize-1] > 52) {
40         printf("J%d\n", List1[MaxSize-1]-52);
41     } else if(List1[MaxSize-1] > 39) {
42         printf("D%d\n", List1[MaxSize-1]-39);
43     } else if(List1[MaxSize-1] > 26) {
44         printf("C%d\n", List1[MaxSize-1]-26);
45     } else if(List1[MaxSize-1] > 13) {
46         printf("H%d\n", List1[MaxSize-1]-13);
47     } else {
48         printf("S%d\n", List1[MaxSize-1]);
49     }
50 
51     return 0;
52 }

 1 #include <cstdio>
 2 
 3 const int N = 54;
 4 char mp[5] = {'S', 'H', 'C', 'D', 'J'};
 5 int start[N], end[N], next[N];
 6 
 7 int main()
 8 {
 9     //freopen("E:\\Temp\\input.txt", "r", stdin);
10 
11     int K;
12     scanf("%d", &K);
13     for(int i=1; i<=N; i++)
14         start[i] = i;
15     for(int i=1; i<=N; i++)
16         scanf("%d", &next[i]);
17 
18     for(int step = 0; step < K; step++) {
19         for(int i=1; i<=N; i++)
20             end[next[i]] = start[i];
21         for(int i=1; i<=N; i++)
22             start[i] = end[i];
23     }
24 
25     for(int i=1; i<=N; i++) {
26         if(i != 1)
27             printf(" ");
28         start[i]--;
29         printf("%c%d", mp[start[i]/13], start[i]%13+1);
30     }
31 
32     return 0;
33 }

 

A1046.Shortest Distance (20)

Description:

The task is really simple: given N exits on a highway which forms a
simple cycle, you are supposed to tell the shortest distance between any
pair of exits.

Input:

Each input file contains one test case. For each case, the first line
contains an integer N (in [3, 105]), followed by N integer
distances D1 D2 … DN, where
Di is the distance between the i-th and the (i+1)-st exits,
and DN is between the N-th and the 1st exits. All the numbers
in a line are separated by a space. The second line gives a positive
integer M (<=104), with M lines follow, each contains a
pair of exit numbers, provided that the exits are numbered from 1 to N.
It is guaranteed that the total round trip distance is no more than
107.

Output:

For each test case, print your results in M lines, each contains the
shortest distance between the corresponding given pair of exits.

Sample Input:

5 1 2 4 14 9
3
1 3
2 5
4 1

Sample Output:

3

10

7

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 const int MAXN = 100005;
 6 int dis[MAXN], A[MAXN];
 7 
 8 int main()
 9 {
10     int sum = 0, query, n, left, right;
11     scanf("%d", &n);
12     for(int i=1; i<=n; i++) {
13         scanf("%d", &A[i]);
14         sum += A[i];
15         dis[i] = sum;
16     }
17     scanf("%d", &query);
18     for(int i=0; i<query; i++) {
19         scanf("%d%d", &left, &right);
20         if(left > right)
21             swap(left, right);
22         int temp = dis[right-1]-dis[left-1];
23         printf("%d\n", min(temp, sum-temp));
24     }
25 
26     return 0;
27 }

 

A1065. A+B and C (64bit) (20)

Description:

Given three integers A, B and C in [-263, 263],
you are supposed to tell whether A+B > C.

Input:

The first line of the input gives the positive number of test cases, T
(<=10). Then T test cases follow, each consists of a single line
containing three integers A, B and C, separated by single spaces.

Ouput:

For each test case, output in one line “Case #X: true” if A+B>C, or
“Case #X: false” otherwise, where X is the case number (starting from
1).

Sample Input:

3
1 2 3
2 3 4
9223372036854775807 -9223372036854775808 0

Output:

Case #1: false
Case #2: true
Case #3: false

 1 #include <cstdio>
 2 
 3 int main()
 4 {
 5     //freopen("E:\\Temp\\input.txt", "r", stdin);
 6 
 7     int T, tcase = 1;
 8     scanf("%d", &T);
 9     while(T--) {
10         long long a, b, c;
11         scanf("%lld%lld%lld", &a, &b, &c);
12         long long res = a+b;
13         bool flag;
14         if(a>0 && b>0 && res<0)
15             flag = true;
16         else if(a<0 && b<0 && res>=0)
17             flag = false;
18         else if(res > c)
19             flag = true;
20         else
21             flag = false;
22         if(flag == true)
23             printf("Case #%d: true\n", tcase++);
24         else
25             printf("Case #%d: false\n", tcase++);
26     }
27 
28     return 0;
29 }

 

B1010. 一元多项式求导 (二五)

Description:

规划函数求1元多项式的导数。(注:xn(n为整数)的一阶导数为n*xn-1。)

Input:

以指数递降方式输入多项式非零项周全和指数(相对值均为不超过1000的整数)。数字间以空格分隔。

Output:

以与输入一样的格式输出导数多项式非零项的全面和指数。数字间以空格分隔,但说起底不可能有结余空格。注意“零多项式”的指数和周全都以0,不过代表为“0
0”。

Sample Input:

3 4 -5 2 6 1 -2 0

Sample Output:

12 3 -10 1 6 0

 1 #include <cstdio>
 2 
 3 int main()
 4 {
 5     //freopen("E:\\Temp\\input.txt", "r", stdin);
 6 
 7     int a[1010] = {0};
 8     int k, e, counter = 0;
 9     while(scanf("%d%d", &k, &e) != EOF) {
10         a[e] = k;
11     }
12     a[0] = 0;
13 
14     for(int i=1; i<=1000; ++i) {
15         a[i-1] = a[i]*i;
16         a[i] = 0;
17         if(a[i-1] != 0)
18             counter++;
19     }
20     if(counter == 0)
21         printf("0 0\n");
22     else {
23         for(int i=1000; i>=0; --i) {
24             if(a[i] != 0) {
25                 printf("%d %d", a[i], i);
26                 counter--;
27                 if(counter != 0)
28                     printf(" ");
29             }
30         }
31     }
32 
33     return 0;
34 }

 

A1002. A+B for Polynomials (25)

Description:

This time, you are supposed to find A+B where A and B are two
polynomials.

Input:

Each input file contains one test case. Each case occupies 2 lines, and
each line contains the information of a polynomial: K N1
aN1 N2 aN2 … NK aNK, where K is the
number of nonzero terms in the polynomial, Ni and aNi (i=1,
2, …, K) are the exponents and coefficients, respectively. It is given
that 1 <= K <= 10,0 <= NK < … < N2 < N1 <=1000.

Output:

For each test case you should output the sum of A and B in one line,
with the same format as the input. Notice that there must be NO extra
space at the end of each line. Please be accurate to 1 decimal place.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 2 1.5 1 2.9 0 3.2

 1 #include <cstdio>
 2 
 3 #define MaxSize 1010
 4 double List[MaxSize];
 5 
 6 int main()
 7 {
 8     //freopen("E:\\Temp\\input.txt", "r", stdin);
 9 
10     int K, expon, counter = 0;
11     double coef;
12     scanf("%d", &K);
13     for(int i=0; i<K; ++i) {
14         scanf("%d %lf", &expon, &coef);
15         List[expon] += coef;
16     }
17     scanf("%d", &K);
18     for(int i=0; i<K; ++i) {
19         scanf("%d %lf", &expon, &coef);
20         List[expon] += coef;
21     }
22 
23     for(int i=0; i<MaxSize; ++i) {
24         if(List[i] != 0)
25             ++counter;
26     }
27     printf("%d", counter);
28     for(int i=MaxSize-1; i>=0; --i) {
29         if(List[i] != 0) {
30             printf(" %d %.1f", i, List[i]);
31         }
32     }
33 
34     return 0;
35 }

 

A1009. Product of Polynomials (25)

Description:

This time, you are supposed to find A*B where A and B are two
polynomials.

Input:

Each input file contains one test case. Each case occupies 2 lines, and
each line contains the information of a polynomial: K N1
aN1轻松模拟习题集。 N2 aN2 … NK aNK, where K is the
number of nonzero terms in the polynomial, Ni and aNi (i=1,
2, …, K) are the exponents and coefficients, respectively. It is given
that 1 <= K <= 10, 0 <= NK < … < N2 < N1 <=1000.

Output:

For each test case you should output the product of A and B in one line,
with the same format as the input. Notice that there must be NO extra
space at the end of each line. Please be accurate up to 1 decimal place.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

3 3 3.6 2 6.0 1 1.6

 1 #include <cstdio>
 2 
 3 #define MaxSize 2010
 4 double List1[MaxSize], List2[MaxSize];
 5 
 6 int main()
 7 {
 8     //freopen("E:\\Temp\\input.txt", "r", stdin);
 9 
10     int K, expon, counter = 0;
11     double coef;
12     scanf("%d", &K);
13     for(int i=0; i<K; ++i) {
14         scanf("%d %lf", &expon, &coef);
15         List1[expon] += coef;
16     }
17     scanf("%d", &K);
18     for(int i=0; i<K; ++i) {
19         scanf("%d %lf", &expon, &coef);
20         for(int j=0; j<MaxSize; j++)
21             List2[expon+j] += List1[j]*coef;
22     }
23 
24     for(int i=0; i<MaxSize; ++i) {
25         if(List2[i] != 0)
26             ++counter;
27     }
28     printf("%d", counter);
29     for(int i=MaxSize-1; i>=0; --i) {
30         if(List2[i] != 0)
31             printf(" %d %.1f", i, List2[i]);
32     }
33 
34     return 0;
35 }

 1 #include <cstdio>
 2 
 3 struct Poly {
 4     int exp;
 5     double cof;
 6 }poly[1001];
 7 double ans[2001];
 8 
 9 int main()
10 {
11     int n, m, number = 0;
12     scanf("%d", &n);
13     for(int i=0; i<n; ++i)
14         scanf("%d %lf", &poly[i].exp, &poly[i].cof);
15     scanf("%d", &m);
16     for(int i=0; i<m; ++i) {
17         int exp;
18         double cof;
19         scanf("%d %lf", &exp, &cof);
20         for(int j=0; j<n; j++)
21             ans[exp+poly[j].exp] += (cof*poly[j].cof);
22     }
23 
24     for(int i=0; i<=2000; ++i) {
25         if(ans[i] != 0)
26             ++number;
27     }
28 
29     printf("%d", number);
30     for(int i=2000; i>=0; --i) {
31         if(ans[i] != 0)
32             printf(" %d %.1f", i, ans[i]);
33     }
34 
35     return 0;
36 }

 

B101八.
榔头剪刀布 (20) Discription:
我们应该都会玩“锤子剪刀布”的游艺:五人还要提交手势,胜…

B101八.
锤子剪刀布 (20)

1042. Shuffling Machine (20)

日子范围

400 ms

内部存款和储蓄器限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Shuffling is a procedure used to randomize a deck of playing cards.
Because standard shuffling techniques are seen as weak, and in order to
avoid “inside jobs” where employees collaborate with gamblers by
performing inadequate shuffles, many casinos employ automatic
shuffling machines
. Your task is to simulate a shuffling machine.

The machine shuffles a deck of 54 cards according to a given random
order and repeats for a given number of times. It is assumed that the
initial status of a card deck is in the following order:

S1, S2, …, S13, H1, H2, …, H13, C1, C2, …, C13, D1, D2, …, D13,
J1, J2

where “S” stands for “Spade”, “H” for “Heart”, “C” for “Club”, “D” for
“Diamond”, and “J” for “Joker”. A given order is a permutation of
distinct integers in [1, 54]. If the number at the i-th position is j,
it means to move the card from position i to position j. For example,
suppose we only have 5 cards: S3, H5, C1, D13 and J2. Given a shuffling
order {4, 2, 5, 3, 1}, the result will be: J2, H5, D13, S3, C1. If we
are to repeat the shuffling again, the result will be: C1, H5, S3, J2,
D13.

Input Specification:

Each input file contains one test case. For each case, the first line
contains a positive integer K (<= 20) which is the number of repeat
times. Then the next line contains the given order. All the numbers in a
line are separated by a space.

Output Specification:

For each test case, print the shuffling results in one line. All the
cards are separated by a space, and there must be no extra space at the
end of the line.

Sample Input:

2
36 52 37 38 3 39 40 53 54 41 11 12 13 42 43 44 2 4 23 24 25 26 27 6 7 8 48 49 50 51 9 10 14 15 16 5 17 18 19 1 20 21 22 28 29 30 31 32 33 34 35 45 46 47

Sample Output:

S7 C11 C10 C12 S1 H7 H8 H9 D8 D9 S11 S12 S13 D10 D11 D12 S3 S4 S6 S10 H1 H2 C13 D2 D3 D4 H6 H3 D13 J1 J2 C1 C2 C3 C4 D1 S5 H5 H11 H12 C6 C7 C8 C9 S2 S8 S9 H10 D5 D6 D7 H4 H13 C5

源代码:

#include <iostream>
#include <cstdio>
using namespace std;
const int N=54;
char mp[5]={'S','H','C','D','J'};
int s[N+1],e[N+1],n[N+1];
int main() {
    int k;
    scanf("%d",&k);
    for(int i=1;i<=N;++i) s[i]=i;
    for(int i=1;i<=N;++i) scanf("%d",&n[i]);
    for(int num=0;num<k;++num) {
        for(int i=1;i<=N;++i) e[n[i]]=s[i];
        for(int i=1;i<=N;++i) s[i]=e[i];
    }
    for(int i=1;i<=N;++i) {
        if(i!=1) printf(" ");
        s[i]--;
        printf("%c%d",mp[s[i]/13],s[i]%13+1);
    }
    return 0;
}

Discription:

Discription:

大家应该都会玩“锤子剪刀布”的玩乐:四人还要提交手势,胜负规则如图所示:

世家应该都会玩“锤子剪刀布”的游乐:三人还要提交手势,胜负规则如图所示:

澳门葡京备用网址 2

澳门葡京备用网址 3

现给出四个人的较量记录,请总计两方的胜、平、负次数,并且付诸双方分别出如何手势的胜算最大。

现给出三个人的交锋记录,请总结两方的胜、平、负次数,并且付诸双方分别出怎么样手势的胜算最大。

Input:

Input:

输入第壹行提交正整数N(<=拾5),即两边交锋的次数。随后N行,每行给出1次比赛的新闻,即甲、乙双方还要提交的的手势。C代表“锤子”、J代表“剪刀”、B代表“布”,第一个假名代表甲方,第3个代表乙方,中间有三个空格。

输入第3行提交正整数N(<=拾5),即双方交锋的次数。随后N行,每行给出三遍交锋的新闻,即甲、乙双方同时提交的的手势。C代表“锤子”、J代表“剪刀”、B代表“布”,第一个字母代表甲方,第1个代表乙方,中间有一个空格。

Output:

Output:

出口第二、2行分别给出甲、乙的胜、平、负次数,数字间以一个空格分隔。第二行提交多个假名,分别代表甲、乙胜球次数最多的手势,中间有叁个空格。假设解不唯一,则输出按字母序最小的解。

出口第壹、二行分别给出甲、乙的胜、平、负次数,数字间以叁个空格分隔。第3行提交三个假名,分别代表甲、乙获胜次数最多的手势,中间有二个空格。要是解不唯1,则输出按字母序最小的解。

Sample
Input:

Sample
Input:

10
C
J
J
B
C
B
B
B
B
C
C
C
C
B
J
B
B
C
J
J

10
C
J
J
B
C
B
B
B
B
C
C
C
C
B
J
B
B
C
J
J

Sample
Output:

Sample
Output:

5 3
2
2 3
5
B
B

5 3
2
2 3
5
B
B

 1 #include <cstdio>
 2 
 3 #define MaxSize 3
 4 
 5 int ListA[MaxSize], ListB[MaxSize], Time[MaxSize];
 6 
 7 int main()
 8 {
 9     //freopen("E:\\Temp\\input.txt", "r", stdin);
10 
11     int N;
12     char c1, c2;
13     scanf("%d", &N);
14     for(int i=0; i<N; i++) {
15         getchar();
16         scanf("%c %c", &c1, &c2);
17             if(c1 == 'B') {
18                 if(c2 == 'B') {
19                     Time[1]++;
20                 } else if(c2 == 'C') {
21                     Time[0]++;
22                     ListA[0]++;
23                 } else {
24                     Time[2]++;
25                     ListB[2]++;
26                 }
27             } else if(c1 == 'C') {
28                 if(c2 == 'B') {
29                     Time[2]++;
30                     ListB[0]++;
31                 } else if(c2 == 'C') {
32                     Time[1]++;
33                 } else {
34                     Time[0]++;
35                     ListA[1]++;
36                 }
37             } else {
38                 if(c2 == 'B') {
39                     Time[0]++;
40                     ListA[2]++;
41                 } else if(c2 == 'C') {
42                     Time[2]++;
43                     ListB[1]++;
44                 } else {
45                     Time[1]++;
46                 }
47             }
48     }
49 
50     printf("%d %d %d\n%d %d %d\n", Time[0], Time[1], Time[2], Time[2], Time[1], Time[0]);
51     if(ListA[0] >= ListA[1]) {
52         if(ListA[0] >= ListA[2])
53             printf("B ");
54         else
55             printf("J ");
56     } else {
57         if(ListA[1] >= ListA[2])
58             printf("C ");
59         else
60             printf("J ");
61     }
62     if(ListB[0] >= ListB[1]) {
63         if(ListB[0] >= ListB[2])
64             printf("B\n");
65         else
66             printf("J\n");
67     } else {
68         if(ListB[1] >= ListB[2])
69             printf("C\n");
70         else
71             printf("J\n");
72     }
73 
74     return 0;
75 }

 1 #include <cstdio>
 2 
 3 int change(char c)
 4 {
 5     if(c == 'B')
 6         return 0;
 7     else if(c == 'C')
 8         return 1;
 9     else
10         return 2;
11 }
12 
13 int main()
14 {
15     //freopen("E:\\Temp\\input.txt", "r", stdin);
16 
17     char mp[3] = {'B', 'C', 'J'};
18     int n;
19     scanf("%d", &n);
20     int times_A[3] = {0}, times_B[3] = {0};
21     int hand_A[3] = {0}, hand_B[3] = {0};
22     char c1, c2;
23     int k1, k2;
24     for(int i=0; i<n; i++) {
25         getchar();
26         scanf("%c %c", &c1, &c2);
27         k1 = change(c1);
28         k2 = change(c2);
29         if((k1+1)%3 == k2) {
30             times_A[0]++;
31             times_B[2]++;
32             hand_A[k1]++;
33         } else if(k1 == k2) {
34             times_A[1]++;
35             times_B[1]++;
36         } else {
37             times_A[2]++;
38             times_B[0]++;
39             hand_B[k2]++;
40         }
41     }
42 
43     printf("%d %d %d\n", times_A[0], times_A[1], times_A[2]);
44     printf("%d %d %d\n", times_B[0], times_B[1], times_B[2]);
45     int id1 = 0, id2 = 0;
46     for(int i=0; i<3; i++) {
47         if(hand_A[i] > hand_A[id1])
48             id1 = i;
49         if(hand_B[i] > hand_B[id2])
50             id2 = i;
51     }
52     printf("%c %c\n", mp[id1], mp[id2]);
53 
54     return 0;
55 }
 1 #include <cstdio>
 2 
 3 #define MaxSize 3
 4 
 5 int ListA[MaxSize], ListB[MaxSize], Time[MaxSize];
 6 
 7 int main()
 8 {
 9     //freopen("E:\\Temp\\input.txt", "r", stdin);
10 
11     int N;
12     char c1, c2;
13     scanf("%d", &N);
14     for(int i=0; i<N; i++) {
15         getchar();
16         scanf("%c %c", &c1, &c2);
17             if(c1 == 'B') {
18                 if(c2 == 'B') {
19                     Time[1]++;
20                 } else if(c2 == 'C') {
21                     Time[0]++;
22                     ListA[0]++;
23                 } else {
24                     Time[2]++;
25                     ListB[2]++;
26                 }
27             } else if(c1 == 'C') {
28                 if(c2 == 'B') {
29                     Time[2]++;
30                     ListB[0]++;
31                 } else if(c2 == 'C') {
32                     Time[1]++;
33                 } else {
34                     Time[0]++;
35                     ListA[1]++;
36                 }
37             } else {
38                 if(c2 == 'B') {
39                     Time[0]++;
40                     ListA[2]++;
41                 } else if(c2 == 'C') {
42                     Time[2]++;
43                     ListB[1]++;
44                 } else {
45                     Time[1]++;
46                 }
47             }
48     }
49 
50     printf("%d %d %d\n%d %d %d\n", Time[0], Time[1], Time[2], Time[2], Time[1], Time[0]);
51     if(ListA[0] >= ListA[1]) {
52         if(ListA[0] >= ListA[2])
53             printf("B ");
54         else
55             printf("J ");
56     } else {
57         if(ListA[1] >= ListA[2])
58             printf("C ");
59         else
60             printf("J ");
61     }
62     if(ListB[0] >= ListB[1]) {
63         if(ListB[0] >= ListB[2])
64             printf("B\n");
65         else
66             printf("J\n");
67     } else {
68         if(ListB[1] >= ListB[2])
69             printf("C\n");
70         else
71             printf("J\n");
72     }
73 
74     return 0;
75 }

 1 #include <cstdio>
 2 
 3 int change(char c)
 4 {
 5     if(c == 'B')
 6         return 0;
 7     else if(c == 'C')
 8         return 1;
 9     else
10         return 2;
11 }
12 
13 int main()
14 {
15     //freopen("E:\\Temp\\input.txt", "r", stdin);
16 
17     char mp[3] = {'B', 'C', 'J'};
18     int n;
19     scanf("%d", &n);
20     int times_A[3] = {0}, times_B[3] = {0};
21     int hand_A[3] = {0}, hand_B[3] = {0};
22     char c1, c2;
23     int k1, k2;
24     for(int i=0; i<n; i++) {
25         getchar();
26         scanf("%c %c", &c1, &c2);
27         k1 = change(c1);
28         k2 = change(c2);
29         if((k1+1)%3 == k2) {
30             times_A[0]++;
31             times_B[2]++;
32             hand_A[k1]++;
33         } else if(k1 == k2) {
34             times_A[1]++;
35             times_B[1]++;
36         } else {
37             times_A[2]++;
38             times_B[0]++;
39             hand_B[k2]++;
40         }
41     }
42 
43     printf("%d %d %d\n", times_A[0], times_A[1], times_A[2]);
44     printf("%d %d %d\n", times_B[0], times_B[1], times_B[2]);
45     int id1 = 0, id2 = 0;
46     for(int i=0; i<3; i++) {
47         if(hand_A[i] > hand_A[id1])
48             id1 = i;
49         if(hand_B[i] > hand_B[id2])
50             id2 = i;
51     }
52     printf("%c %c\n", mp[id1], mp[id2]);
53 
54     return 0;
55 }

 

 

A1042.
Shuffling Machine (20)

A1042.
Shuffling Machine (20)

Description:

Description:

Shuffling
is a procedure used to randomize a deck of playing cards. Because
standard shuffling techniques are seen as weak, and in order to avoid
“inside jobs” where employees collaborate with gamblers by performing
inadequate shuffles, many casinos employ automatic shuffling machines.
Your task is to simulate a shuffling machine.

Shuffling
is a procedure used to randomize a deck of playing cards. Because
standard shuffling techniques are seen as weak, and in order to avoid
“inside jobs” where employees collaborate with gamblers by performing
inadequate shuffles, many casinos employ automatic shuffling machines.
Your task is to simulate a shuffling machine.

The
machine shuffles a deck of 54 cards according to a given random order
and repeats for a given number of times. It is assumed that the initial
status of a card deck is in the following order:

The
machine shuffles a deck of 54 cards according to a given random order
and repeats for a given number of times. It is assumed that the initial
status of a card deck is in the following order:

S1,
S2, …, S13, H1, H2, …, H13, C1, C2, …, C13, D1, D2, …, D13, J1,
J2

S1,
S2, …, S13, H1, H2, …, H13, C1, C2, …, C13, D1, D2, …, D13, J1,
J2

where
“S” stands for “Spade”, “H” for “Heart”, “C” for “Club”, “D” for
“Diamond”, and “J” for “Joker”. A given order is a permutation of
distinct integers in [1, 54]. If the number at the i-th position is j,
it means to move the card from position i to position j. For example,
suppose we only have 5 cards: S3, H5, C1, D13 and J2. Given a shuffling
order {4, 2, 5, 3, 1}, the result will be: J2, H5, D13, S3, C1. If we
are to repeat the shuffling again, the result will be: C1, H5, S3, J2,
D13.

where
“S” stands for “Spade”, “H” for “Heart”, “C” for “Club”, “D” for
“Diamond”, and “J” for “Joker”. A given order is a permutation of
distinct integers in [1, 54]. If the number at the i-th position is j,
it means to move the card from position i to position j. For example,
suppose we only have 5 cards: S3, H5, C1, D13 and J2. Given a shuffling
order {4, 2, 5, 3, 1}, the result will be: J2, H5, D13, S3, C1. If we
are to repeat the shuffling again, the result will be: C1, H5, S3, J2,
D13.

Input:

Input:

Each
input file contains one test case. For each case, the first line
contains a positive integer K (<= 20) which is the number of repeat
times. Then the next line contains the given order. All the numbers in a
line are separated by a space.

Each
input file contains one test case. For each case, the first line
contains a positive integer K (<= 20) which is the number of repeat
times. Then the next line contains the given order. All the numbers in a
line are separated by a space.

Output:

Output:

For
each test case, print the shuffling results in one line. All the cards
are separated by a space, and there must be no extra space at the end of
the line.

For
each test case, print the shuffling results in one line. All the cards
are separated by a space, and there must be no extra space at the end of
the line.

Sample
Input:

Sample
Input:

2
36 52
37 38 3 39 40 53 54 41 11 12 13 42 43 44 2 4 23 24 25 26 27 6 7 8 48 49
50 51 9 10 14 15 16 5 17 18 19 1 20 21 22 28 29 30 31 32 33 34 35 45 46
47

2
36 52
37 38 3 39 40 53 54 41 11 12 13 42 43 44 2 4 23 24 25 26 27 6 7 8 48 49
50 51 9 10 14 15 16 5 17 18 19 1 20 21 22 28 29 30 31 32 33 34 35 45 46
47

Sample
Output:

Sample
Output:

S7 C11
C10 C12 S1 H7 H8 H9 D8 D9 S11 S12 S13 D10 D11 D12 S3 S4 S6 S10 H1 H2 C13
D2 D3 D4 H6 H3 D13 J1 J2 C1 C2 C3 C4 D1 S5 H5 H11 H12 C6 C7 C8 C9 S2 S8
S9 H10 D5 D6 D7 H4 H13 C5

S7 C11
C10 C12 S1 H7 H8 H9 D8 D9 S11 S12 S13 D10 D11 D12 S3 S4 S6 S10 H1 H2 C13
D2 D3 D4 H6 H3 D13 J1 J2 C1 C2 C3 C4 D1 S5 H5 H11 H12 C6 C7 C8 C9 S2 S8
S9 H10 D5 D6 D7 H4 H13 C5

 1 #include <cstdio>
 2 
 3 #define MaxSize 55
 4 
 5 int List1[MaxSize], List2[MaxSize], List3[MaxSize];
 6 
 7 int main()
 8 {
 9     //freopen("E:\\Temp\\input.txt", "r", stdin);
10 
11     int N, temp;
12     scanf("%d", &N);
13     for(int i=1; i<MaxSize; i++) {
14         List1[i] = i;
15         scanf("%d", &List2[i]);
16     }
17     for(int i=0; i<N; i++) {
18         for(int j=1; j<MaxSize; j++) {
19             List3[List2[j]] = List1[j];
20         }
21         for(int k=1; k<MaxSize; k++) {
22             List1[k] = List3[k];
23         }
24     }
25 
26     for(int i=1; i<MaxSize-1; i++) {
27         if(List1[i] > 52) {
28             printf("J%d ", List1[i]-52);
29         } else if(List1[i] > 39) {
30             printf("D%d ", List1[i]-39);
31         } else if(List1[i] > 26) {
32             printf("C%d ", List1[i]-26);
33         } else if(List1[i] > 13) {
34             printf("H%d ", List1[i]-13);
35         } else {
36             printf("S%d ", List1[i]);
37         }
38     }
39     if(List1[MaxSize-1] > 52) {
40         printf("J%d\n", List1[MaxSize-1]-52);
41     } else if(List1[MaxSize-1] > 39) {
42         printf("D%d\n", List1[MaxSize-1]-39);
43     } else if(List1[MaxSize-1] > 26) {
44         printf("C%d\n", List1[MaxSize-1]-26);
45     } else if(List1[MaxSize-1] > 13) {
46         printf("H%d\n", List1[MaxSize-1]-13);
47     } else {
48         printf("S%d\n", List1[MaxSize-1]);
49     }
50 
51     return 0;
52 }

 1 #include <cstdio>
 2 
 3 const int N = 54;
 4 char mp[5] = {'S', 'H', 'C', 'D', 'J'};
 5 int start[N], end[N], next[N];
 6 
 7 int main()
 8 {
 9     //freopen("E:\\Temp\\input.txt", "r", stdin);
10 
11     int K;
12     scanf("%d", &K);
13     for(int i=1; i<=N; i++)
14         start[i] = i;
15     for(int i=1; i<=N; i++)
16         scanf("%d", &next[i]);
17 
18     for(int step = 0; step < K; step++) {
19         for(int i=1; i<=N; i++)
20             end[next[i]] = start[i];
21         for(int i=1; i<=N; i++)
22             start[i] = end[i];
23     }
24 
25     for(int i=1; i<=N; i++) {
26         if(i != 1)
27             printf(" ");
28         start[i]--;
29         printf("%c%d", mp[start[i]/13], start[i]%13+1);
30     }
31 
32     return 0;
33 }
 1 #include <cstdio>
 2 
 3 #define MaxSize 55
 4 
 5 int List1[MaxSize], List2[MaxSize], List3[MaxSize];
 6 
 7 int main()
 8 {
 9     //freopen("E:\\Temp\\input.txt", "r", stdin);
10 
11     int N, temp;
12     scanf("%d", &N);
13     for(int i=1; i<MaxSize; i++) {
14         List1[i] = i;
15         scanf("%d", &List2[i]);
16     }
17     for(int i=0; i<N; i++) {
18         for(int j=1; j<MaxSize; j++) {
19             List3[List2[j]] = List1[j];
20         }
21         for(int k=1; k<MaxSize; k++) {
22             List1[k] = List3[k];
23         }
24     }
25 
26     for(int i=1; i<MaxSize-1; i++) {
27         if(List1[i] > 52) {
28             printf("J%d ", List1[i]-52);
29         } else if(List1[i] > 39) {
30             printf("D%d ", List1[i]-39);
31         } else if(List1[i] > 26) {
32             printf("C%d ", List1[i]-26);
33         } else if(List1[i] > 13) {
34             printf("H%d ", List1[i]-13);
35         } else {
36             printf("S%d ", List1[i]);
37         }
38     }
39     if(List1[MaxSize-1] > 52) {
40         printf("J%d\n", List1[MaxSize-1]-52);
41     } else if(List1[MaxSize-1] > 39) {
42         printf("D%d\n", List1[MaxSize-1]-39);
43     } else if(List1[MaxSize-1] > 26) {
44         printf("C%d\n", List1[MaxSize-1]-26);
45     } else if(List1[MaxSize-1] > 13) {
46         printf("H%d\n", List1[MaxSize-1]-13);
47     } else {
48         printf("S%d\n", List1[MaxSize-1]);
49     }
50 
51     return 0;
52 }

 1 #include <cstdio>
 2 
 3 const int N = 54;
 4 char mp[5] = {'S', 'H', 'C', 'D', 'J'};
 5 int start[N], end[N], next[N];
 6 
 7 int main()
 8 {
 9     //freopen("E:\\Temp\\input.txt", "r", stdin);
10 
11     int K;
12     scanf("%d", &K);
13     for(int i=1; i<=N; i++)
14         start[i] = i;
15     for(int i=1; i<=N; i++)
16         scanf("%d", &next[i]);
17 
18     for(int step = 0; step < K; step++) {
19         for(int i=1; i<=N; i++)
20             end[next[i]] = start[i];
21         for(int i=1; i<=N; i++)
22             start[i] = end[i];
23     }
24 
25     for(int i=1; i<=N; i++) {
26         if(i != 1)
27             printf(" ");
28         start[i]--;
29         printf("%c%d", mp[start[i]/13], start[i]%13+1);
30     }
31 
32     return 0;
33 }

 

 

A1046.Shortest
Distance (20)

A1046.Shortest
Distance (20)

Description:

Description:

The
task is really simple: given N exits on a highway which forms a simple
cycle, you are supposed to tell the shortest distance between any pair
of exits.

The
task is really simple: given N exits on a highway which forms a simple
cycle, you are supposed to tell the shortest distance between any pair
of exits.

Input:

Input:

Each
input file contains one test case. For each case, the first line
contains an integer N (in [澳门葡京备用网址 ,3, 105]), followed by N integer
distances D1 D2 … DN, where
Di is the distance between the i-th and the (i+1)-st exits,
and DN is between the N-th and the 1st exits. All the numbers
in a line are separated by a space. The second line gives a positive
integer M (<=104), with M lines follow, each contains a
pair of exit numbers, provided that the exits are numbered from 1 to N.
It is guaranteed that the total round trip distance is no more than
107.

Each
input file contains one test case. For each case, the first line
contains an integer N (in [3, 105]), followed by N integer
distances D1 D2 … DN, where
Di is the distance between the i-th and the (i+1)-st exits,
and DN is between the N-th and the 1st exits. All the numbers
in a line are separated by a space. The second line gives a positive
integer M (<=104), with M lines follow, each contains a
pair of exit numbers, provided that the exits are numbered from 1 to N.
It is guaranteed that the total round trip distance is no more than
107.

Output:

Output:

For
each test case, print your results in M lines, each contains the
shortest distance between the corresponding given pair of exits.

For
each test case, print your results in M lines, each contains the
shortest distance between the corresponding given pair of exits.

Sample
Input:

Sample
Input:

5 1 2
4 14 9
3
1
3
2
5
4
1

5 1 2
4 14 9
3
1
3
2
5
4
1

Sample
Output:

Sample
Output:

3

3

10

10

7

7

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 const int MAXN = 100005;
 6 int dis[MAXN], A[MAXN];
 7 
 8 int main()
 9 {
10     int sum = 0, query, n, left, right;
11     scanf("%d", &n);
12     for(int i=1; i<=n; i++) {
13         scanf("%d", &A[i]);
14         sum += A[i];
15         dis[i] = sum;
16     }
17     scanf("%d", &query);
18     for(int i=0; i<query; i++) {
19         scanf("%d%d", &left, &right);
20         if(left > right)
21             swap(left, right);
22         int temp = dis[right-1]-dis[left-1];
23         printf("%d\n", min(temp, sum-temp));
24     }
25 
26     return 0;
27 }
 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 
 5 const int MAXN = 100005;
 6 int dis[MAXN], A[MAXN];
 7 
 8 int main()
 9 {
10     int sum = 0, query, n, left, right;
11     scanf("%d", &n);
12     for(int i=1; i<=n; i++) {
13         scanf("%d", &A[i]);
14         sum += A[i];
15         dis[i] = sum;
16     }
17     scanf("%d", &query);
18     for(int i=0; i<query; i++) {
19         scanf("%d%d", &left, &right);
20         if(left > right)
21             swap(left, right);
22         int temp = dis[right-1]-dis[left-1];
23         printf("%d\n", min(temp, sum-temp));
24     }
25 
26     return 0;
27 }

 

 

A1065.
A+B and C (64bit) (20)

A1065.
A+B and C (64bit) (20)

Description:

Description:

Given
three integers A, B and C in [-263, 263], you
are supposed to tell whether A+B > C.

Given
three integers A, B and C in [-263, 263], you
are supposed to tell whether A+B > C.

Input:

Input:

The
first line of the input gives the positive number of test cases, T
(<=10). Then T test cases follow, each consists of a single line
containing three integers A, B and C, separated by single spaces.

The
first line of the input gives the positive number of test cases, T
(<=10). Then T test cases follow, each consists of a single line
containing three integers A, B and C, separated by single spaces.

Ouput:

Ouput:

For
each test case, output in one line “Case #X: true” if A+B>C, or
“Case #X: false” otherwise, where X is the case number (starting from
1).

For
each test case, output in one line “Case #X: true” if A+B>C, or
“Case #X: false” otherwise, where X is the case number (starting from
1).

Sample
Input:

Sample
Input:

3
1 2
3
2 3
4
9223372036854775807
-9223372036854775808 0

3
1 2
3
2 3
4
9223372036854775807
-9223372036854775808 0

Output:

Output:

Case
#1: false
Case
#2: true
Case
#3: false

Case
#1: false
Case
#2: true
Case
#3: false

 1 #include <cstdio>
 2 
 3 int main()
 4 {
 5     //freopen("E:\\Temp\\input.txt", "r", stdin);
 6 
 7     int T, tcase = 1;
 8     scanf("%d", &T);
 9     while(T--) {
10         long long a, b, c;
11         scanf("%lld%lld%lld", &a, &b, &c);
12         long long res = a+b;
13         bool flag;
14         if(a>0 && b>0 && res<0)
15             flag = true;
16         else if(a<0 && b<0 && res>=0)
17             flag = false;
18         else if(res > c)
19             flag = true;
20         else
21             flag = false;
22         if(flag == true)
23             printf("Case #%d: true\n", tcase++);
24         else
25             printf("Case #%d: false\n", tcase++);
26     }
27 
28     return 0;
29 }
 1 #include <cstdio>
 2 
 3 int main()
 4 {
 5     //freopen("E:\\Temp\\input.txt", "r", stdin);
 6 
 7     int T, tcase = 1;
 8     scanf("%d", &T);
 9     while(T--) {
10         long long a, b, c;
11         scanf("%lld%lld%lld", &a, &b, &c);
12         long long res = a+b;
13         bool flag;
14         if(a>0 && b>0 && res<0)
15             flag = true;
16         else if(a<0 && b<0 && res>=0)
17             flag = false;
18         else if(res > c)
19             flag = true;
20         else
21             flag = false;
22         if(flag == true)
23             printf("Case #%d: true\n", tcase++);
24         else
25             printf("Case #%d: false\n", tcase++);
26     }
27 
28     return 0;
29 }

 

 

B10十.
一元多项式求导 (二伍)

B10十.
一元多项式求导 (2伍)

Description:

Description:

规划函数求1元多项式的导数。(注:xn(n为整数)的1阶导数为n*xn-1。)

统一筹划函数求一元多项式的导数。(注:xn(n为整数)的一阶导数为n*xn-1。)

Input:

Input:

以指数递降格局输入多项式非零项周全和指数(相对值均为不超越一千的平头)。数字间以空格分隔。

以指数递降形式输入多项式非零项周到和指数(绝对值均为不抢先一千的整数)。数字间以空格分隔。

Output:

Output:

以与输入一样的格式输出导数多项式非零项的周密和指数。数字间以空格分隔,但最后无法有多余空格。注意“零多项式”的指数和周详都以0,然则代表为“0
0”。

以与输入同样的格式输出导数多项式非零项的周密和指数。数字间以空格分隔,但最终不能够有剩余空格。注意“零多项式”的指数和周到都以0,可是表示为“0
0”。

Sample
Input:

Sample
Input:

3 4 -5
2 6 1 -2 0

3 4 -5
2 6 1 -2 0

Sample
Output:

Sample
Output:

12 3
-10 1 6 0

12 3
-10 1 6 0

 1 #include <cstdio>
 2 
 3 int main()
 4 {
 5     //freopen("E:\\Temp\\input.txt", "r", stdin);
 6 
 7     int a[1010] = {0};
 8     int k, e, counter = 0;
 9     while(scanf("%d%d", &k, &e) != EOF) {
10         a[e] = k;
11     }
12     a[0] = 0;
13 
14     for(int i=1; i<=1000; ++i) {
15         a[i-1] = a[i]*i;
16         a[i] = 0;
17         if(a[i-1] != 0)
18             counter++;
19     }
20     if(counter == 0)
21         printf("0 0\n");
22     else {
23         for(int i=1000; i>=0; --i) {
24             if(a[i] != 0) {
25                 printf("%d %d", a[i], i);
26                 counter--;
27                 if(counter != 0)
28                     printf(" ");
29             }
30         }
31     }
32 
33     return 0;
34 }
 1 #include <cstdio>
 2 
 3 int main()
 4 {
 5     //freopen("E:\\Temp\\input.txt", "r", stdin);
 6 
 7     int a[1010] = {0};
 8     int k, e, counter = 0;
 9     while(scanf("%d%d", &k, &e) != EOF) {
10         a[e] = k;
11     }
12     a[0] = 0;
13 
14     for(int i=1; i<=1000; ++i) {
15         a[i-1] = a[i]*i;
16         a[i] = 0;
17         if(a[i-1] != 0)
18             counter++;
19     }
20     if(counter == 0)
21         printf("0 0\n");
22     else {
23         for(int i=1000; i>=0; --i) {
24             if(a[i] != 0) {
25                 printf("%d %d", a[i], i);
26                 counter--;
27                 if(counter != 0)
28                     printf(" ");
29             }
30         }
31     }
32 
33     return 0;
34 }

 

 

A1002.
A+B for Polynomials (25)

A1002.
A+B for Polynomials (25)

Description:

Description:

This
time, you are supposed to find A+B where A and B are two
polynomials.

This
time, you are supposed to find A+B where A and B are two
polynomials.

Input:

Input:

Each
input file contains one test case. Each case occupies 2 lines, and each
line contains the information of a polynomial: K N1 aN1 N2
aN2 … NK aNK, where K is the number of nonzero
terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the
exponents and coefficients, respectively. It is given that 1 <= K
<= 10,0 <= NK < … < N2 < N1 <=1000.

Each
input file contains one test case. Each case occupies 2 lines, and each
line contains the information of a polynomial: K N1 aN1 N2
aN2 … NK aNK, where K is the number of nonzero
terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the
exponents and coefficients, respectively. It is given that 1 <= K
<= 10,0 <= NK < … < N2 < N1 <=1000.

Output:

Output:

For
each test case you should output the sum of A and B in one line, with
the same format as the input. Notice that there must be NO extra space
at the end of each line. Please be accurate to 1 decimal place.

For
each test case you should output the sum of A and B in one line, with
the same format as the input. Notice that there must be NO extra space
at the end of each line. Please be accurate to 1 decimal place.

Sample
Input:

Sample
Input:

2 1
2.4 0 3.2
2 2
1.5 1 0.5

2 1
2.4 0 3.2
2 2
1.5 1 0.5

Sample
Output:

Sample
Output:

3 2
1.5 1 2.9 0 3.2

3 2
1.5 1 2.9 0 3.2

 1 #include <cstdio>
 2 
 3 #define MaxSize 1010
 4 double List[MaxSize];
 5 
 6 int main()
 7 {
 8     //freopen("E:\\Temp\\input.txt", "r", stdin);
 9 
10     int K, expon, counter = 0;
11     double coef;
12     scanf("%d", &K);
13     for(int i=0; i<K; ++i) {
14         scanf("%d %lf", &expon, &coef);
15         List[expon] += coef;
16     }
17     scanf("%d", &K);
18     for(int i=0; i<K; ++i) {
19         scanf("%d %lf", &expon, &coef);
20         List[expon] += coef;
21     }
22 
23     for(int i=0; i<MaxSize; ++i) {
24         if(List[i] != 0)
25             ++counter;
26     }
27     printf("%d", counter);
28     for(int i=MaxSize-1; i>=0; --i) {
29         if(List[i] != 0) {
30             printf(" %d %.1f", i, List[i]);
31         }
32     }
33 
34     return 0;
35 }
 1 #include <cstdio>
 2 
 3 #define MaxSize 1010
 4 double List[MaxSize];
 5 
 6 int main()
 7 {
 8     //freopen("E:\\Temp\\input.txt", "r", stdin);
 9 
10     int K, expon, counter = 0;
11     double coef;
12     scanf("%d", &K);
13     for(int i=0; i<K; ++i) {
14         scanf("%d %lf", &expon, &coef);
15         List[expon] += coef;
16     }
17     scanf("%d", &K);
18     for(int i=0; i<K; ++i) {
19         scanf("%d %lf", &expon, &coef);
20         List[expon] += coef;
21     }
22 
23     for(int i=0; i<MaxSize; ++i) {
24         if(List[i] != 0)
25             ++counter;
26     }
27     printf("%d", counter);
28     for(int i=MaxSize-1; i>=0; --i) {
29         if(List[i] != 0) {
30             printf(" %d %.1f", i, List[i]);
31         }
32     }
33 
34     return 0;
35 }

 

 

A1009.
Product of Polynomials (25)

A1009.
Product of Polynomials (25)

Description:

Description:

This
time, you are supposed to find A*B where A and B are two
polynomials.

This
time, you are supposed to find A*B where A and B are two
polynomials.

Input:

Input:

Each
input file contains one test case. Each case occupies 2 lines, and each
line contains the information of a polynomial: K N1 aN1 N2
aN2 … NK aNK, where K is the number of nonzero
terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the
exponents and coefficients, respectively. It is given that 1 <= K
<= 10, 0 <= NK < … < N2 < N1 <=1000.

Each
input file contains one test case. Each case occupies 2 lines, and each
line contains the information of a polynomial: K N1 aN1 N2
aN2 … NK aNK, where K is the number of nonzero
terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the
exponents and coefficients, respectively. It is given that 1 <= K
<= 10, 0 <= NK < … < N2 < N1 <=1000.

Output:

Output:

For
each test case you should output the product of A and B in one line,
with the same format as the input. Notice that there must be NO extra
space at the end of each line. Please be accurate up to 1 decimal
place.

For
each test case you should output the product of A and B in one line,
with the same format as the input. Notice that there must be NO extra
space at the end of each line. Please be accurate up to 1 decimal
place.

Sample
Input:

Sample
Input:

2 1
2.4 0 3.2
2 2
1.5 1 0.5

2 1
2.4 0 3.2
2 2
1.5 1 0.5

Sample
Output:

Sample
Output:

3 3
3.6 2 6.0 1 1.6

3 3
3.6 2 6.0 1 1.6

 1 #include <cstdio>
 2 
 3 #define MaxSize 2010
 4 double List1[MaxSize], List2[MaxSize];
 5 
 6 int main()
 7 {
 8     //freopen("E:\\Temp\\input.txt", "r", stdin);
 9 
10     int K, expon, counter = 0;
11     double coef;
12     scanf("%d", &K);
13     for(int i=0; i<K; ++i) {
14         scanf("%d %lf", &expon, &coef);
15         List1[expon] += coef;
16     }
17     scanf("%d", &K);
18     for(int i=0; i<K; ++i) {
19         scanf("%d %lf", &expon, &coef);
20         for(int j=0; j<MaxSize; j++)
21             List2[expon+j] += List1[j]*coef;
22     }
23 
24     for(int i=0; i<MaxSize; ++i) {
25         if(List2[i] != 0)
26             ++counter;
27     }
28     printf("%d", counter);
29     for(int i=MaxSize-1; i>=0; --i) {
30         if(List2[i] != 0)
31             printf(" %d %.1f", i, List2[i]);
32     }
33 
34     return 0;
35 }

 1 #include <cstdio>
 2 
 3 struct Poly {
 4     int exp;
 5     double cof;
 6 }poly[1001];
 7 double ans[2001];
 8 
 9 int main()
10 {
11     int n, m, number = 0;
12     scanf("%d", &n);
13     for(int i=0; i<n; ++i)
14         scanf("%d %lf", &poly[i].exp, &poly[i].cof);
15     scanf("%d", &m);
16     for(int i=0; i<m; ++i) {
17         int exp;
18         double cof;
19         scanf("%d %lf", &exp, &cof);
20         for(int j=0; j<n; j++)
21             ans[exp+poly[j].exp] += (cof*poly[j].cof);
22     }
23 
24     for(int i=0; i<=2000; ++i) {
25         if(ans[i] != 0)
26             ++number;
27     }
28 
29     printf("%d", number);
30     for(int i=2000; i>=0; --i) {
31         if(ans[i] != 0)
32             printf(" %d %.1f", i, ans[i]);
33     }
34 
35     return 0;
36 }
 1 #include <cstdio>
 2 
 3 #define MaxSize 2010
 4 double List1[MaxSize], List2[MaxSize];
 5 
 6 int main()
 7 {
 8     //freopen("E:\\Temp\\input.txt", "r", stdin);
 9 
10     int K, expon, counter = 0;
11     double coef;
12     scanf("%d", &K);
13     for(int i=0; i<K; ++i) {
14         scanf("%d %lf", &expon, &coef);
15         List1[expon] += coef;
16     }
17     scanf("%d", &K);
18     for(int i=0; i<K; ++i) {
19         scanf("%d %lf", &expon, &coef);
20         for(int j=0; j<MaxSize; j++)
21             List2[expon+j] += List1[j]*coef;
22     }
23 
24     for(int i=0; i<MaxSize; ++i) {
25         if(List2[i] != 0)
26             ++counter;
27     }
28     printf("%d", counter);
29     for(int i=MaxSize-1; i>=0; --i) {
30         if(List2[i] != 0)
31             printf(" %d %.1f", i, List2[i]);
32     }
33 
34     return 0;
35 }

 1 #include <cstdio>
 2 
 3 struct Poly {
 4     int exp;
 5     double cof;
 6 }poly[1001];
 7 double ans[2001];
 8 
 9 int main()
10 {
11     int n, m, number = 0;
12     scanf("%d", &n);
13     for(int i=0; i<n; ++i)
14         scanf("%d %lf", &poly[i].exp, &poly[i].cof);
15     scanf("%d", &m);
16     for(int i=0; i<m; ++i) {
17         int exp;
18         double cof;
19         scanf("%d %lf", &exp, &cof);
20         for(int j=0; j<n; j++)
21             ans[exp+poly[j].exp] += (cof*poly[j].cof);
22     }
23 
24     for(int i=0; i<=2000; ++i) {
25         if(ans[i] != 0)
26             ++number;
27     }
28 
29     printf("%d", number);
30     for(int i=2000; i>=0; --i) {
31         if(ans[i] != 0)
32             printf(" %d %.1f", i, ans[i]);
33     }
34 
35     return 0;
36 }

 

 

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