lightoj 1030 Discovering Gold

1030 - Discovering Gold

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Time Limit: 2 second(s)

Memory Limit: 32 MB

You are in a cave, a long cave! The cave can be represented by 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 you throw a perfect 6 sided dice. If you get X in the dice after throwing, you add X to your position and collect all the gold from the new
position. If your new position is outside the cave, then you keep throwing again until you get a suitable result. When you reach the Nth position you stop your journey. Now you are given the information about the cave, you have
to find out the expected number of gold you can collect using 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 line contains N space separated integers. The ith integer of this line denotes the amount
of gold you will get if you come to the ith cell. You may safely assume that all the given integers will be non-negative and no integer will be greater than 1000.

Output

For each case, print the case number and the expected number of gold you will collect. Errors less than 10-6 will be ignored.

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

计算数学期望，用dp数组保存到达该点的概率，那么该点的数学期望就是概率乘上该点的金矿数；

#include<iostream>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<stack>
#include<queue>
#include<iomanip>
#include<map>
#include<set>
#include<functional>
#define pi 3.14159265358979323846
using namespace std;
int cave[105];
double dp[105];
int main()
{
int T,N,cnt=0;
scanf("%d",&T);
while(T--)
{
scanf(" %d",&N);
memset(dp,0,sizeof(dp));
dp[0]=1;
for(int i=0;i<N;++i)
{
scanf("%d",&cave[i]);
}
double sum=0;
for(int i=0;i<N;++i)
{
if(i==0||i==N-1) sum+=cave[i];
else sum+=cave[i]*dp[i];
int temp=1;
while(i+temp<N-1)
{
++temp;
}
if(temp>6)
temp=6;
for(int j=1;j<=temp;++j)
{
dp[i+j]+=1.0/(temp*1.0)*dp[i];
}
}
printf("Case %d: %.10lf\n",++cnt,sum);
}
return 0;
}