uva 10759(数论)

来源：互联网发布：质谱软件编辑：程序博客网时间：2024/06/03 20:53

题意：有n个骰子，给出一个目标值x，求出得到的所有骰子点数之和大于x的概率。

题解：分母肯定是6^n，分子需要dp得到，开一个二维数组f[i][j]表示i个骰子组成和大于等于j的情况有多少种。状态转移方程是f[i][j + k] += f[i - 1][j]，然后两个值求最大公约数约分一下就可以了。

#include <stdio.h>#include <string.h>#include <math.h>#define ll long longint n, sum;ll f[25][160];ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);}void init() {memset(f, 0, sizeof(f));f[0][0] = 1;for (int i = 1; i <= 24; i++)for (int j = 0; j <= 150; j++)if (f[i - 1][j])for (int k = 1; k <= 6; k++)f[i][j + k] += f[i - 1][j];}int main() {init();while (scanf("%d%d", &n, &sum) && n + sum) {if (sum <= n) {printf("1\n");continue;}ll res1 = 0, res2 = pow(6, n);for (int i = sum; i <= 150; i++)res1 += f[n][i];if (res1 == 0) {printf("0\n");continue;}long long temp = gcd(res1, res2);printf("%lld/%lld\n", res1 / temp, res2 / temp);}return 0;}

0 0