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]
= 1;

nu = de - dfs(m, 0);
tmp = gcd(de, nu);
if (nu == 0)
printf("0\n");
else if(nu == de)
printf("1\n");
else
printf("%lld/%lld\n", nu / tmp, de / tmp);
}
return 0;
}