lightoj1030(概率dp)

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1030

You are in a cave, a long cave! The cave can be representedby a 1 x N grid. Each cell of the cave can contain any amount of gold.

Initially you are in position 1. Now each turn youthrow a perfect 6 sided dice. If you get X in the dice afterthrowing, you add X to your position and collect all the gold from thenew position. If your new position is outside the cave, then you keep throwingagain until you get a suitable result. When you reach the N^thposition you stop your journey. Now you are given the information about thecave, you have to find out the expected number of gold you can collectusing the given procedure.

Input

Input starts with an integer T (≤ 100),denoting the number of test cases.

Each case contains a blank line and an integer N (1≤ N ≤ 100) denoting the dimension of the cave. The next linecontains N space separated integers. The i^th integerof this line denotes the amount of gold you will get if you come to the i^thcell. You may safely assume that all the given integers will be non-negativeand no integer will be greater than 1000.

Output

For each case, print the case number and the expected numberof gold you will collect. Errors less than 10^-6 will beignored.

Sample Input	Output for Sample Input
3 1 101 2 10 3 3 3 6 9	Case 1: 101.0000000000 Case 2: 13.000 Case 3: 15

思路：本来不会，看的别人的题解才写出来。直接参考这位大牛的吧。。。

https://blog.csdn.net/L123012013048/article/details/46361807

贴代码：

#include<bits/stdc++.h>
using namespace std;
int v[110];
double dp[110];
int main()
{
	int n,t;
	scanf("%d",&t);
	int cnt=0;
	while(t--)
	{
		scanf("%d",&n);
		cnt++;
		memset(v,0,sizeof(v));
		for(int i=0;i<=n;i++)
		dp[i]=0;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&v[i]);
		}
		dp[n]=v[n];
		for(int i=n-1;i>=1;i--)
		{
			for(int j=1;j<=min(n-i,6);j++)
			{
				dp[i]=dp[i]+1.0/min(n-i,6)*dp[i+j];
			}
			dp[i]+=v[i];
		}
		printf("Case %d: %.10lf\n",cnt,dp[1]);
	}
	return 0;
}