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:

#include <iostream>
using namespace std;

typedef long long LL;
const int MOD = 1e9 + 7;
LL c[50][25];

LL powMod(int a, int b, int c) {
    LL res = 1;
    while (b) {
        if (b & 1) res = (res * a) % c;
        a = (LL)a * a % c;
        b >>= 1;
    }
    return res;
}

void init() {
    c[0][0] = 1;
    for (int i = 1; i < 50; i++) {
        c[i][0] = 1; c[i - 1][i] = 0;
        for (int j = 1; j <= i; j++) {
            c[i][j] = ((LL)c[i - 1][j] + c[i - 1][j - 1]) % MOD;
        }
    }
}

int main(void) {
    init();
    int n, m;
    while (cin >> n >> m) {
        if (n > m) swap(n, m);
        if (m < 25 || m == 25 && m - n < 2) {cout << "0" << endl; continue;}
        if (m > 25 && m - n != 2) {cout << "0" << endl; continue;}
        if (m == 25) {
            cout << c[24 + n][n] << endl;
        } else {
            cout << ((LL)c[48][24] * powMod(2, n - 24, MOD)) % MOD << endl;
        }
    }
    return 0;
}