题目链接

Tatyana is a big sports fan and she likes volleyball a lot! She writes down the final scores of the game after it has ended in her notebook.

If you are not familiar with the rules of volleyball, here‘s a brief:

• 2 teams play in total
• During the course of the game, each team gets points, and thus increases its score by 1.
• The initial score is 0 for both teams.

The game ends when

• One of the teams gets 25 points and another team has < 24 points ( strictly less than 24).
• If the score ties at 24:24, the teams continue to play until the absolute difference between the scores is 2.

Given the final score of a game in the format A:*B* i.e., the first team has scored A points and the second has scored B points, can you print the number of different sequences of getting points by teams that leads to this final score?

Input Format
The first line contains A and the second line contains B.

Constraints

0 ≤ A , B ≤ 109

Output Format
Output the number of different sequences of getting points by the teams that leads to the final score A : B. Final means that the game should be over after this score is reached. If the number is larger than 109+7, output number modulo 109 + 7. Print 0 if no such volleyball game ends with the given score.

Example input #00

325

Example output #00

2925

Example input #01

2417

Example output #01

0

Explanation #01

There‘s no game of volleyball that ends with a score of 24 : 17.

题目大意：给出一局排球比赛的比分，求有多少种方式得到这个比分。。

知识点：组合数学

Accepted Code:

 1 #include <iostream> 2 using namespace std; 3  4 typedef long long LL; 5 const int MOD = 1e9 + 7; 6 LL c[50][25]; 7  8 LL powMod(int a, int b, int c) { 9     LL res = 1;10     while (b) {11         if (b & 1) res = (res * a) % c;12         a = (LL)a * a % c;13         b >>= 1;14     }15     return res;16 }17 18 void init() {19     c[0][0] = 1;20     for (int i = 1; i < 50; i++) {21         c[i][0] = 1; c[i - 1][i] = 0;22         for (int j = 1; j <= i; j++) {23             c[i][j] = ((LL)c[i - 1][j] + c[i - 1][j - 1]) % MOD;24         }25     }26 }27 28 int main(void) {29     init();30     int n, m;31     while (cin >> n >> m) {32         if (n > m) swap(n, m);33         if (m < 25 || m == 25 && m - n < 2) {cout << "0" << endl; continue;}34         if (m > 25 && m - n != 2) {cout << "0" << endl; continue;}35         if (m == 25) {36             cout << c[24 + n][n] << endl;37         } else {38             cout << ((LL)c[48][24] * powMod(2, n - 24, MOD)) % MOD << endl;39         }40     }41     return 0;42 }

Hackerrank--Volleyball Match