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1027 - A Dangerous Maze

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Time Limit: 2 second(s)

Memory Limit: 32 MB

You are in a maze; seeing n doors in front of you in beginning. You can choose any door you like. The probability for choosing a door is equal for all doors.

If you choose the i^th door, it can either take you back to the same position where you begun in x_i minutes, or can take you out of the maze after x_i minutes. If you come back to the same position, you can't remember anything. So, every time you come to the beginning position, you have no past experience.

Now you want to find the expected time to get out of the maze.

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 number of doors. The next line contains n space separated integers. If the i^th integer (x_i) is positive, you can assume that the i^th door will take you out of maze after x_i minutes. If it's negative, then the i^th door will take you back to the beginning position after abs(x_i) minutes. You can safely assume that 1 ≤ abs(x_i) ≤ 10000.

Output

For each case, print the case number and the expected time to get out of the maze. If it's impossible to get out of the maze, print 'inf'. Print the result in p/q format. Where p is the numerator of the result andq is the denominator of the result and they are relatively prime. See the samples for details.

Sample Input	Output for Sample Input
3   1 1   2 -10 -3   3 3 -6 -9	Case 1: 1/1 Case 2: inf Case 3: 18/1

 

題意：T組數據，每組數據n個門，每個門後都有一個數字ti，如果爲正則經過ti走出迷宮，反之，經過時間ti後回到原位，求走出迷宮的期望

解析：設當前走出迷宮時間的期望E

E = p1 * sum1 + p2 * (sum2 + E) 其中sum1爲選擇爲正的門所花費的期望時間，sum2表示選擇爲負的門所花費的時間。如果選擇了負的門，則又要面臨相同的選擇。p1表示選擇正的門的概率，p2表示選擇負的門的概率

對該式子進行整理後得到 E = （sum1 + sum2） / p

代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <string>
#include <algorithm>
#include <list>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdlib>
using namespace std;

int main()
{
	//freopen("in.txt","r",stdin);
	int T,a;
	scanf("%d",&T);
	for(int t = 1; t <= T; t ++)
    {
        printf("Case %d: ",t);
        int n,p = 0;
        int sum1 = 0,sum2 = 0;
        scanf("%d",&n);
        for(int i = 0; i < n; i ++)
        {
            scanf("%d",&a);
            if(a> 0)
            {
                sum1 += a;
                p ++;
            }
            else
            {
                sum2 += -1 * a;
            }
        }
        int fenzi = sum1 + sum2;
        int fenmu = p;
        if(p == 0)
        {
            printf("inf\n");
            continue;
        }
        else
        {
            int k = __gcd(fenzi,fenmu);
            printf("%d/%d\n",fenzi/k,fenmu/k);
        }
    }
	return 0;
}