ZOJ 3329 One Person Game(概率DP、求期望）

题目链接：点击打开链接One Person GameTime Limit: 1 Second Memory Limit: 32768 KB Special JudgeThere 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:Set the counter to 0 at first.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.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.InputThere 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 testcase 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).OutputFor 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, PengSource: The 7th Zhejiang Provincial Collegiate Programming ContestOne Person GameTime Limit: 1 Second Memory Limit: 32768 KB Special JudgeThere 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:Set the counter to 0 at first.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.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.InputThere 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 testcase 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).OutputFor 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, PengSource: The 7th Zhejiang Provincial Collegiate Programming Contest题意：有三个色子，分别有k1,k2,k3面，投掷这三个色子，如果他们面朝上的数分别是a,b,c，分数归为0，否则加上这三个数的和，分数超过n，游戏 结束，否则继续投色子，求投掷色子次数的期望。起始分数为0。思路：设dp[i]为从当前数位i达到目标状态需要投色子的次数，pk为投掷k分的概率，p0为回到0的概率则dp[i]=∑(pk*dp[i+k])+dp[0]*p0+1;都和dp[0]有关系，而且dp[0]就是我们所求，为常数

设dp[i]=A[i]*dp[0]+B[i];
代入上述方程右边得到：
dp[i]=∑(pk*A[i+k]*dp[0]+pk*B[i+k])+dp[0]*p0+1
=(∑(pk*A[i+k])+p0)dp[0]+∑(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[0]和B[0].
那么  dp[0]=B[0]/(1-A[0]);

代码：

#include<stdio.h>#include<string.h>int main(){int t,n,k1,k2,k3,a,b,c,i,j,h;double dp[800],A[800],B[800],p[100];scanf("%d",&t);while(t--){scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);memset(p,0,sizeof(p));double p0=1.0/(k1*k2*k3);for(i=1;i<=k1;i++)for(j=1;j<=k2;j++)for(h=1;h<=k3;h++)if(i!=a||j!=b||h!=c)p[i+j+h]+=1.0/(k1*k2*k3);memset(A,0,sizeof(A));memset(B,0,sizeof(B));for(i=n;i>=0;i--){A[i]=p0;B[i]=1;for(j=1;j<=k1+k2+k3;j++){A[i]+=p[j]*A[i+j];B[i]+=p[j]*B[i+j];}}printf("%.16lf\n",B[0]/(1-A[0]));}}