light oj A Dangerous Maze（概率dp求期望）

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

dp[i]表示从点i出发能到得到金子数量的期望

枚举每次走的步数

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 10010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int x = 0;
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}

double dp[1010];
double a[1010];

int main()
{
//    fread;
    int tc;
    int cs=1;
    scanf("%d",&tc);
    while(tc--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%lf",&a[i]);
        MEM(dp,0);
        dp
=a
;
        for(int i=n-1;i>=1;i--)
        {
            dp[i]=a[i];
            int step=min(6,n-i);
            for(int j=1;j<=step;j++)
                dp[i]+=1.0/(step*1.0)*dp[i+j];
        }
        printf("Case %d: %.6lf\n",cs++,dp[1]);
    }
    return 0;
}