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
参考 bin神 的博客，如果自己做绝对做不出，数学不好真是硬伤
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<vector>#include<set>#include<map>#define L(x) (x<<1)#define R(x) (x<<1|1)#define MID(x,y) ((x+y)>>1)#define eps 1e-8//typedef __int64 ll;#define fre(i,a,b) for(i = a; i <b; i++)#define free(i,b,a) for(i = b; i >= a;i–)#define mem(t, v) memset ((t) , v, sizeof(t))#define ssf(n)scanf("%s", n)#define sf(n)scanf("%d", &n)#define sff(a,b) scanf("%d %d", &a, &b)#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)#define pfprintf#define bugpf("Hi\n")using namespace std;#define INF 0x3f3f3f3f#define N 1005double p[N],dp[N];double A[N],B[N];int a,b,c,n,k1,k2,k3;double pp;int main(){int i,j,s;int t;sf(t);while(t–){sf(n);sfff(k1,k2,k3);sfff(a,b,c);mem(p,0);pp=1.0/k1/k2/k3;fre(i,1,k1+1) fre(j,1,k2+1)fre(s,1,k3+1)if(i!=a||j!=b||s!=c)p[i+j+s]+=pp;mem(A,0);mem(B,0);free(i,n,0){A[i]=pp;B[i]=1;fre(j,3,k1+k2+k3+1){A[i]+=p[j]*A[i+j];B[i]+=p[j]*B[i+j];}}double ans=B[0]/(1-A[0]);pf("%.15f\n",ans);}return 0;}
，此刻睡觉的口水将变成明天流下的泪水。
