poj1218 THE DRUNK JAILER

最新推荐文章于 2019-05-14 15:01:00 发布

转载最新推荐文章于 2019-05-14 15:01:00 发布 · 153 阅读

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原文链接：http://www.cnblogs.com/w0w0/archive/2011/11/21/2256810.html

本文介绍了一个经典的算法问题——醉酒狱卒问题。问题描述了一位狱卒在喝醉后进行的一系列操作，最终确定哪些牢房被解锁，允许囚犯逃脱。通过给出的C++代码示例，演示了如何通过编程解决这个问题。

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THE DRUNK JAILER

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17539		Accepted: 11167

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

#include<iostream>
using namespace std;
int main()
{
int num;
int sum;
int state[101];
int n;
    cin>>num;
while(num--)
    {
int i,j;
        cin>>n;
        sum=0;
for(i=1;i<=n;i++)
        {
            state[i]=1;
        }
for(i=1;i<=n;i++)
        {
for(j=1;j<=n;j++)
            {
if(j%i==0)
                    state[j]=!state[j];
            }
        }
for(i=1;i<=n;i++)
        {
if(state[i]==0)
                sum++;
        }
        cout<<sum<<endl;
    }
return 0;
}

转载于:https://www.cnblogs.com/w0w0/archive/2011/11/21/2256810.html