## ZOJ 3329 One Person Game（概率DP，求期望）

One Person Game

Time Limit: 1 Second      Memory Limit: 32768 KB      Special Judge

There is a very simple and interesting one-person game. You have 3 dice, namely Die1, Die2 and Die3. Die1 has K1 faces. Die2 has K2 faces. Die3 has K3 faces. All the dice are fair dice, so the probability of rolling each value, 1 to K1, K2, K3 is exactly 1 / K1, 1 / K2 and 1 / K3. You have a counter, and the game is played as follow:

1. Set the counter to 0 at first.
2. Roll the 3 dice simultaneously. If the up-facing number of Die1 is a, the up-facing number of Die2 is b and the up-facing number of Die3 is c, set the counter to 0. Otherwise, add the counter by the total value of the 3 up-facing numbers.
3. If the counter's number is still not greater than n, go to step 2. Otherwise the game is ended.

Calculate the expectation of the number of times that you cast dice before the end of the game.

Input

There are multiple test cases. The first line of input is an integer T (0 < T <= 300) indicating the number of test cases. Then T test cases follow. Each test case is a line contains 7 non-negative integers n, K1, K2, K3, a, b, c (0 <= n <= 500, 1 < K1, K2, K3 <= 6, 1 <= a <= K1, 1 <= b <= K2, 1 <= c <= K3).

Output

For each test case, output the answer in a single line. A relative error of 1e-8 will be accepted.

Sample Input

2
0 2 2 2 1 1 1
0 6 6 6 1 1 1


Sample Output

1.142857142857143
1.004651162790698


Author: CAO, Peng
Source: The 7th Zhejiang Provincial Collegiate Programming Contest

/*
ZOJ 3329

dp[i]=∑(pk*A[i+k]*dp+pk*B[i+k])+dp*p0+1
=(∑(pk*A[i+k])+p0)dp+∑(pk*B[i+k])+1;
明显A[i]=(∑(pk*A[i+k])+p0)
B[i]=∑(pk*B[i+k])+1
先递推求得A和B.
那么  dp=B/(1-A);
*/
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;

double A,B;
double p;
int main()
{
int T;
int k1,k2,k3,a,b,c;
int n;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);
double p0=1.0/k1/k2/k3;
memset(p,0,sizeof(p));
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]+=p0;
memset(A,0,sizeof(A));
memset(B,0,sizeof(B));
for(int i=n;i>=0;i--)
{
A[i]=p0;B[i]=1;
for(int j=1;j<=k1+k2+k3;j++)
{
A[i]+=A[i+j]*p[j];
B[i]+=B[i+j]*p[j];
}
}
printf("%.16lf\n",B/(1-A));
}
return 0;
}

posted on 2012-10-03 00:22  kuangbin  阅读(3205)  评论(3编辑  收藏

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