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Light OJ 1030

編輯：關於C++

1030 - Discovering Gold

PDF (English) Statistics Forum 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 N^th 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 i^th integer of this line denotes the amount of gold you will get if you come to the i^th 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

101

10 3

3 6 9

Case 1: 101.0000000000

Case 2: 13.000

Case 3: 15

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include


#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 = 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); //這個地方很重要，因為如果超出了n就是無效的
	  pf("Case %d: %.6lf\n",++ca,dp[1]);
	}
   return 0;
}