THE DRUNK JAILER 分类： POJ 2015-06-10 14:50 13人阅读 评论(0) 收藏

THE DRUNK JAILER

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24918		Accepted: 15632

Description

A certain prison contains a long hall of n cells, each right next to each other. Each cell has a prisoner in it, and each cell is locked.

One night, the jailer gets bored and decides to play a game. For round 1 of the game, he takes a drink of whiskey,and then runs down the hall unlocking each cell. For round 2, he takes a drink of whiskey, and then runs down the

hall locking every other cell (cells 2, 4, 6, ?). For round 3, he takes a drink of whiskey, and then runs down the hall. He visits every third cell (cells 3, 6, 9, ?). If the cell is locked, he unlocks it; if it is unlocked, he locks it. He

repeats this for n rounds, takes a final drink, and passes out.

Some number of prisoners, possibly zero, realizes that their cells are unlocked and the jailer is incapacitated. They immediately escape.

Given the number of cells, determine how many prisoners escape jail.

Input

The first line of input contains a single positive integer. This is the number of lines that follow. Each of the following lines contains a single integer between 5 and 100, inclusive, which is the number of cells n.

Output

For each line, you must print out the number of prisoners that escape when the prison has n cells.

Sample Input

2
5
100

Sample Output

2
10
poj的一道水题，就是n个牢房，犯人从第一个牢房开始每次改变牢房的状态，如果是锁的，就打开，没有锁，就锁上；
循环n次，看看能逃出去几个犯人
#include <map>
#include <set>
#include <string>
#include <cctype>
#include <climits>
#include <vector>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#define INF 0x3f3f3f3f
#define ClearAll(A,T) memset(A,T,sizeof(A))

using namespace std;

const int Max=100;
int a[110];
int main()
{

    int n,m;
    while(~scanf("%d",&n))
    {
        if(n==0)
        {
            continue;
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&m);
            int sum=0;
            memset(a,0,sizeof(a));
            for(int j=2;j<=m;j++)
            {
                for(int k=j;k<=m;k+=j)
                {
                    a[k]=a[k]==0?1:0;
                }
            }
            for(int i=1;i<=m;i++)
            {
                if(!a[i])
                sum++;
            }
            printf("%d\n",sum);
        }
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。