| 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 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.
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 |
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
#define eps 1e-8
typedef __int64 ll;
#define fre(i,a,b) for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define ssf(n) scanf("%s", n)
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define bug pf("Hi\n")
using namespace std;
#define INF 0x3f3f3f3f
#define N 10005
double dp[N];
int main()
{
int i,j,t,n,ca=0;
sf(t);
while(t--)
{
mem(dp,0);
sf(n);
fre(i,1,n+1)
scanf("%lf",&dp[i]);
for(i=n-1;i>=1;i--)
for(j=1;j<=6;j++)
dp[i]+=dp[i+j]/min(6,n-i); //这个地方很重要，因为如果超出了n就是无效的
pf("Case %d: %.6lf\n",++ca,dp[1]);
}
return 0;
}
Light OJ 1030 - Discovering Gold（期望）
原文：http://blog.csdn.net/u014737310/article/details/45334237