UVA - 10759 Dice Throwing

题目大意：给出n和x，计算在丢n个色子，出现的点数大于等于x的概率，要求分式最简。

解题思路：num[i][j] 与num[i][j + 1]中间的增长个数与用i个色子丢出的点数为j的情况有关，然而求用i个色子丢出点数的情况则是非常好求的：num[i][j]  = ∑(j + 1 ≤ k ≤ j + 6)num[i][k].

然后就可以进一步的去求丢出i个色子丢出的点数小于x的情况。

#include <cstdio>#include <cstring>long long dp[160][30];long long dfs(long long m, int c) {if (dp[m][c] != -1) return dp[m][c];dp[m][c] = 0;for (int i =1 ; i <= 6; i++)if (m - i > 0)dp[m][c] += dfs(m - i, c + 1);return dp[m][c];}long long gcd(long long a, long long b) {return b == 0 ? a : gcd(b, a % b);}int main() {long long n, m, nu, tmp;while (scanf("%lld%lld", &n, &m), n) {long long de = 1;for (int i = 1; i <= n; i++)de *= 6;memset(dp, -1, sizeof(dp));for (int i = 0; i <= m; i++)dp[i][n] = 1;nu = de - dfs(m, 0);tmp = gcd(de, nu);if (nu == 0)printf("0\n");else if(nu == de)printf("1\n");elseprintf("%lld/%lld\n", nu / tmp, de / tmp);}return 0;}