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:
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
我们可以推出公式dp[i]=sum{pk*dp[i+k]}+p0*dp[0]+1,dp[i]表示当前分数是i是时的到游戏结束的期望,那么dp[0],就是要求的,其实,我们可以发现一个归律,求期望都是从后住前推的,比如,这里,是由i+k推到i的,求概率的时候是刚好相反的,但我们发现求出的这个式子,是个环形的,所以要变形,因为,每一个都是和dp[0]相关的,我们可以设,dp[i]=a[i]*dp[0]+b[i],那么dp[0]不就是,b[i]/(1-a[i])了么,我们,把这个式子代入上式就可以得到,dp[i]=sum{pk*a[i+k]}+p0,b[i]=sum{pk*b[i+k]}+1,这样,就可以马上求结果来了!
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; #define MAXN 550 double pa[MAXN],pb[MAXN],dp[MAXN],p[MAXN]; int main() { int tcase,i,j,k,n,a,b,c,k1,k2,k3; double p0; scanf("%d",&tcase); while(tcase--) { scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c); p0=1.0/k1/k2/k3; memset(pa,0,sizeof(pa)); memset(pb,0,sizeof(pb)); memset(dp,0,sizeof(dp)); memset(p,0,sizeof(p)); for(i=1;i<=k1;i++) for(j=1;j<=k2;j++) for(k=1;k<=k3;k++) if(i!=a||j!=b||k!=c)//不同时相等 p[i+j+k]+=p0; int temp=k1+k2+k3; for(i=n;i>=0;i--) { pa[i]=p0; pb[i]=1; for(k=1;k<=temp;k++) { pa[i]+=pa[i+k]*p[k]; pb[i]+=pb[i+k]*p[k]; } } printf("%.15f\n",pb[0]/(1.0-pa[0])); } return 0; }