Hackerrank--Volleyball Match

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

3 25 

Example output #00

2925

Example input #01

24 17 

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 }