【暴力】HRBUST 1011 THE DRUNK JAILER

最新推荐文章于 2018-05-27 01:41:53 发布

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本文探讨了一个有趣的算法问题——醉酒狱卒游戏。通过分析狱卒如何在不同轮次中开关监狱大门的过程，实现了一个简单的C语言程序来计算最终能够成功逃脱的囚犯数量。

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THE DRUNK JAILER

Time Limit: 1000 MS

Memory Limit: 65536 K

Total Submit: 626(462 users)

Total Accepted: 498(450 users)

Rating:

Special Judge: No

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

Sample Output

2
10

题目意思不难理解，有点像灯泡问题，有n个监狱，第一次打开全部的门，第二次遇到2的倍数的监狱的时候如果门开着就关上，如果关上就打开...以此类推，到第n次的时候，所有是n的被数的门当中，如果门的上一个状态是打开的就关上，关上的就打开，问你最后有多少监狱的门是打开的。

数据很水，直接暴力。

附上AC代码

#include<stdio.h>
#include<string.h>

int a[106];

void solve(void)
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        for(int i = 1 ; i <= n ; i++)
        {
            for(int j = 1 ; j <= n ; j++)
            {
                if(j%i==0)
                {
                    if(a[j]==0) a[j] = 1;
                    else a[j] = 0;
                }
            }
        }
        int ans = 0;
        for(int i = 1 ; i <= n ; i++) if(a[i]==1) ans++;
        printf("%d\n",ans);
    }
}


int main(void)
{
    solve();
    return 0;
}