UVA10759 - Dice Throwing(dp+gcd)
题目大意:n个色子,求n个色子之和不小于x的概率。
解题思路:因为可以将题目转化成求n个色子和小于x的数目,最后再6^n减去就这个数目就是大于等于x的数目了。因为n最大就24,这样还是可以用long long来存放,最后输出要求的是分数形式,将分子分母用gcd约分一下输出即可。
代码:
#include <cstdio>
#include <cstring>
typedef long long ll;
const int maxn = 25;
const int maxm = 155;
int N, X;
ll f[maxn][maxm], t[maxn];
void init () {
t[0] = 1LL;
for (int i = 1; i < maxn; i++)
t[i] = t[i - 1] * 6;
}
ll gcd (ll a, ll b) {
return b == 0? a: gcd(b, a%b);
}
ll dp (int n, int x) {
ll& ans = f[n][x];
if (ans != -1)
return ans;
if (n == N) {
if (x < X)
return ans = 1;
return ans = 0;
}
ans = 0;
for (int i = 1; i <= 6; i++)
if (x + i < X)
ans += dp(n + 1, x + i);
else
break;
return ans;
}
int main () {
init();
while (scanf ("%d%d", &N, &X) && (N||X)) {
memset(f, -1, sizeof (f));
ll a = t[N] - dp(0, 0);
if (a == 0) {
printf ("0\n");
continue;
}
if (a == t[N]) {
printf ("1\n");
continue;
}
ll tmp = gcd(a, t[N]);
printf ("%lld/%lld\n", a/tmp, t[N]/tmp);
}
}
时间: 2024-11-05 02:29:04