题意：有三个骰子，分别有k1,k2,k3个面。
每次掷骰子，如果三个面分别为a,b,c则分数置0，否则加上三个骰子的分数之和。
当分数大于n时结束。求游戏的期望步数。初始分数为0
设dp[i]表示达到i分时到达目标状态(即i = n)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];
即为dp[i + k] = A[i + k] * dp[0] + B[i + k];
将其代入原式：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;
所以可得：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]);
Source Code:
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0) using namespace std; typedef long long ll ;
typedef unsigned long long ull ;
typedef unsigned int uint ;
typedef unsigned char uchar ; template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;} const double eps = 1e- ;
const int N = ;
const int M = * ;
const ll P = 10000000097ll ;
const int MAXN = ; int n, k1, k2, k3, ka, kb, kc;
double a[], b[], p[]; int main(){
std::ios::sync_with_stdio(false);
int i, j, t, k, u, v, numCase = ;
cin >> t;
while(t--){
cin >> n >> k1 >> k2 >> k3 >> ka >> kb >> kc;
memset(p, , sizeof(p));
for(i = ; i <= k1; ++i){
for(j = ; j <= k2; ++j){
for(k = ; k <= k3; ++k){
if(i == ka && j == kb && k == kc){
p[] += 1.0 / (k1 * k2 * k3);
continue;
}
p[i + j + k] += 1.0 / (k1 * k2 * k3);
}
}
}
memset(a, , sizeof(a));
memset(b, , sizeof(b));
for(i = n; i >= ; --i){
a[i] = p[];
b[i] = 1.0;
for(k = ; k <= k1 + k2 + k3; ++k){
a[i] += p[k] * a[i + k];
b[i] += p[k] * b[i + k];
}
}
printf("%.16f\n",b[] / (1.0 - a[]));
} return ;
}