总结模拟习题集

Sample
Output:

大家应该都会玩“锤子剪刀布”的娱乐:五个人还要提交手势,胜负规则如图所示:

世家应该都会玩“锤子剪刀布”的游戏:两个人同时提交手势,胜负规则如图所示:

Input:

 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 }

Sample
Input:

Output:

 

A1046.Shortest
Distance (20)

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.

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.

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, 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
Output:

Output:

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.

 

Description:

Sample
Output:

 

3 3
3.6 2 6.0 1 1.6

3 2
1.5 1 2.9 0 3.2

3 4 -5
2 6 1 -2 0

Input:

Input:

Sample
Input:

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

以指数递降方式输入多项式非零项周详和指数(相对值均为不当先一千的平头)。数字间以空格分隔。

 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 }

5 3
2
2 3
5
B
B

Sample
Input:

Input:

A1009.
Product of Polynomials (25)

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

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

Output:

7

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.

Sample
Input:

Sample
Input:

 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 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 }

A1042.
Shuffling Machine (20)

Description:

Description:

Sample
Input:

以与输入相同的格式输出导数多项式非零项的周详和指数。数字间以空格分隔,但结尾不恐怕有多余空格。注意“零多项式”的指数和周详都以0,不过代表为“0
0”。

B1018.
榔头剪刀布 (20)

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

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:

2 1
2.4 0 3.2
2 2
1.5 1 0.5

7

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

Output:

10

 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 }

Description:

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

Input:

 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 }

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:

3 2
1.5 1 2.9 0 3.2

 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 }

Input:

2 1
2.4 0 3.2
2 2
1.5 1 0.5

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

B1010.
一元多项式求导 (25)

B1010.
一元多项式求导 (25)

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.

出口第1 、2行分别给出甲、乙的胜、平、负次数,数字间以2个空格分隔。第叁行提交三个字母,分别代表甲、乙赢球次数最多的手势,中间有三个空格。若是解不唯一,则输出按字母序最小的解。

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.

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.

Description:

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

Description:

Sample
Input:

Output:

以指数递降方式输入多项式非零项周全和指数(相对值均为不超越一千的整数)。数字间以空格分隔。

输入第③行提交正整数N(<=105),即双方交锋的次数。随后N行,每行给出一回交锋的新闻,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”,第一个假名代表甲方,第3个代表乙方,中间有贰个空格。

Sample
Input:

A1042.
Shuffling Machine (20)

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
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.

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.

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.

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.

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.

Output:

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

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

2 1
2.4 0 3.2
2 2
1.5 1 0.5

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.

输入第叁行提交正整数N(<=105),即两边交锋的次数。随后N行,每行给出一回竞技的新闻,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”,第三个字母代表甲方,第四个代表乙方,中间有3个空格。

Sample
Output:

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

3 3
3.6 2 6.0 1 1.6

 

Sample
Input:

 

Input:

3
1 2
3
2 3
4
9223372036854775807
-9223372036854775808 0

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

Description:

A1002.
A+B for Polynomials (25)

Output:

Output:

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:

 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 }

Description:

现给出多少人的交锋记录,请总结双方的胜、平、负次数,并且付诸双方各自出什么手势的胜算最大。

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.

 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 }

12 3
-10 1 6 0

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

B1018.
锤子剪刀布 (20)

Sample
Output:

 

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.

 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 }

3 4 -5
2 6 1 -2 0

现给出多少人的较量记录,请总结双方的胜、平、负次数,并且付诸双方各自出什么样手势的胜算最大。

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:

Input:

A1002.
A+B for Polynomials (25)

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

Sample
Input:

2 1
2.4 0 3.2
2 2
1.5 1 0.5

Description:

Sample
Input:

Sample
Output:

3
1 2
3
2 3
4
9223372036854775807
-9223372036854775808 0

 

Ouput:

A1046.Shortest
Distance (20)

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

图片 1

Description:

Input:

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.

 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 }

Input:

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.

 

Output:

输出第② 、2行分别给出甲、乙的胜、平、负次数,数字间以三个空格分隔。第叁行提交多少个假名,分别代表甲、乙赢球次数最多的手势,中间有3个空格。假诺解不唯一,则输出按字母序最小的解。

Sample
Output:

 

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).

 

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.

 

 

Sample
Input:

Sample
Output:

Description:

Description:

 

 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 }

图片 2

 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 }

以与输入相同的格式输出导数多项式非零项的周到和指数。数字间以空格分隔,但说到底不大概有多余空格。注意“零多项式”的指数和周到都以0,不过代表为“0
0”。

Sample
Input:

10

Sample
Output:

Discription:

Output:

安顿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:

3

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

Discription:

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

5 3
2
2 3
5
B
B

Sample
Output:

Ouput:

12 3
-10 1 6 0

 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 }

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

Sample
Output:

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

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

Output:

 

规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。)

A1009.
Product of Polynomials (25)

Output:

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).

Input:

3

Sample
Output: