杭电2053，简单易懂的C语言代码！！！

Re:题意：有无限多的电灯排成一列，一开始都是关，操作无限多次，第i次操作会把编号为i和i的倍数的电灯改变状态。问最后第i盏电灯的状态是开还是关    //后边有数据。。。

#include<stdio.h>
int main()
{
    int b,n,i;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==1)//第一次操作，开关全变为1；
        printf("1\n");
        else
        {
            b=0;
            for(i=2;i<=n;i++)//从2开始判断约数，开关从1开始；
            {
                if(n%i==0)
                b++;
            }
            if(b%2==0)//判断开关变化次数，（1 0 1 0 1 0........);
            printf("1\n");
            else
            printf("0\n");
        }
    }
    return 0;
} 

Consider the second test case:

The initial condition : 0 0 0 0 0 …

After the first operation : 1 1 1 1 1 …

After the second operation : 1 0 1 0 1 …

After the third operation : 1 0 0 0 1 …

After the fourth operation : 1 0 0 1 1 …

After the fifth operation : 1 0 0 1 0 …

The later operations cannot change the condition of the fifth lamp any more. So the answer is 0.