题目链接：?problemId=3754
There is a very simple and interesting one-person game. You have 3 dice, namelyDie1, Die2 and Die3. Die1 hasK1 faces. Die2 has K2 faces.Die3 has K3 faces. All the dice are fair dice, so the probability of rolling each value, 1 toK1, 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 isa, the up-facing number of Die2 is b and the up-facing number ofDie3 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.
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 integersn, 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
20 2 2 2 1 1 10 6 6 6 1 1 1
Sample Output
1.1428571428571431.004651162790698
PS:
代码如下：#include <cstdio>#include <cstring>int main(){int t;int n;int k1, k2 ,k3;int a, b, c;double p[47];scanf("%d",&t);while(t–){memset(p,0,sizeof(p));scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);double p0 = 1.0/(k1*k2*k3);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;}}}}double a[700], b[700];memset(a, 0,sizeof(a));memset(b, 0,sizeof(b));for(int i = n; i >= 0; i–){for(int k = 3; k <= k1+k2+k3; k++){a[i] += a[i+k] * p[k];b[i] += b[i+k] * p[k];}a[i]+=p0;b[i]+=1;}printf("%.15lf\n",b[0]/(1-a[0]));}return 0;}
，带着感恩的心启程，学会爱，爱父母，爱自己，爱朋友，爱他人。
