lightoj 1030 - Discovering Gold 【期望】

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 Nthposition 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 ithinteger 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

 

PROBLEM SETTER: JANE ALAM JAN

题意：给定n个格子以及每个格子上的gold值，你从第一个格子出发，每次掷1-6的骰子，根据掷出的值前进。若当前位置 + 掷出的值 > n，则重新掷骰子，到达第n个格子游戏结束。问你获得gold值的期望。

思路：设dp[i]为到达i个格子获得gold值的期望，则有dp[i] = (dp[i+1]/6 + ... + dp[i+6]/6) + dp[i]。
初始化dp[i] = 第i个格子的gold值。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (3000+10)
#define MAXM (50000000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
int gcd(int a, int b){
return b == 0 ? a : gcd(b, a%b);
}
double dp[110], num[110];
int main()
{
int t, kcase = 1; Ri(t);
W(t)
{
int n; Ri(n);
for(int i = 0; i < n; i++)
Rf(num[i]), dp[i] = num[i];
for(int i = n-2; i >= 0; i--)
{
int choose = min(6, n - i - 1);
for(int j = 1; j <= choose; j++)
dp[i] += dp[i+j] / (double)choose;
}
printf("Case %d: %.7lf\n", kcase++, dp[0]);
}
return 0;
}