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UVA10759 - Dice Throwing(dp+gcd)

UVA10759 - Dice Throwing(dp+gcd)

题目链接

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);

    }
}

UVA10759 - Dice Throwing(dp+gcd)