题干：
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:
- 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.
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 1Sample Output
1.142857142857143
1.004651162790698题目大意：有三个均匀的骰子，分别有k1,k2,k3个面分别代表点数1~k1,1~k2,1~k3。初始分数是0。每一回合掷一波骰子，当掷三个骰子的点数分别为a,b,c的时候，分数清零，否则分数加上三个骰子的点数和，当分数>n的时候结束。求需要掷骰子的次数的期望。
解题报告：
自己讲的也不好，也是参考一个题解吧：（链接）
写出的公式是有后续性的。我们需要弄一个递推公式，消去后续性。
本题通过代换系数，化简后求系数。一般形成环的用高斯消元法求解。但是此题都是和dp[0]相关。所有可以分离出系数。设dp[i]表示达到i分时到达目标状态的期望，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]);AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 1000 + 5;
ll ans;
int k1,k2,k3,a,b,c,n;
int ci[666];
double A[MAX],B[MAX],dp[MAX];//dp[i]代表当前分数为i时，结束游戏所要掷骰子的次数的期望值
int main()
{int t;cin>>t;while(t--) {scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);memset(ci,0,sizeof ci);for(int i = 0; i<MAX; i++) A[i]=B[i]=0;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) continue;ci[i+j+k]++;}}}double p0=1.0/(k1*k2*k3);for(int i = n; i>=0; i--) {for(int j = 3; j<=k1+k2+k3; j++) {A[i] += ci[j]*1.0/(k1*k2*k3) * A[i+j];B[i] += ci[j]*1.0/(k1*k2*k3) * B[i+j];}A[i] += p0;B[i] += 1;}printf("%.12f\n",B[0]/(1-A[0]));}return 0 ;
}总结：
刚开始WA了半天，发现是初始化的时候直接把ci和A数组放到一起用for初始化了，但是其实这样是不行的，因为你MAX=1000，但是ci只开到了666，所以就越界了呀。
