ZOJ-3329-One Person Game

ACM模版

描述

题解

kuangbin 大佬的题解，神乎其技：

代码

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAXN = 600;int n, k1, k2, k3, a, b, c;double A[MAXN], B[MAXN], P[MAXN];int main(int argc, const char * argv[]){    int T;    cin >> T;    while (T--)    {        memset(P, 0, sizeof(P));        memset(A, 0, sizeof(A));        memset(B, 0, sizeof(B));        scanf("%d%d%d%d%d%d%d", &n, &k1, &k2, &k3, &a, &b, &c);        P[0] = 1.0 / k1 / k2 / k3; //  回到零的概率        for (int i = 1; i <= k1; i++)        {            for (int j = 1; j <= k2; j++)            {                for (int k = 1; k <= k3; k++)                {                    if (i != a || j != b || k != c)                    {                        P[i + j + k] += P[0];   //  投掷出 i j k 的概率等于 P[0]                    }                }            }        }        for (int i = n; i >= 0; i--)        {            A[i] = P[0];            B[i] = 1;            int t = k1 + k2 + k3;            for (int j = 1; j <= t; j++)            {                A[i] += A[i + j] * P[j];                B[i] += B[i + j] * P[j];            }        }        printf("%.15lf\n", B[0] / (1 - A[0]));    }    return 0;}