2053 Switch Game

最新推荐文章于 2025-09-08 21:57:50 发布

原创最新推荐文章于 2025-09-08 21:57:50 发布 · 465 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#杭电ACM #c++ #c

c++ 同时被 3 个专栏收录

286 篇文章

订阅专栏

285 篇文章

订阅专栏

杭电ACM

103 篇文章

订阅专栏

Switch Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10457 Accepted Submission(s): 6347

Problem Description

There are many lamps in a line. All of them are off at first. A series of operations are carried out on these lamps. On the i-th operation, the lamps whose numbers are the multiple of i change the condition ( on to off and off to on ).

Input

Each test case contains only a number n ( 0< n<= 10^5) in a line.

Output

Output the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).

Sample Input

1
5

Sample Output

1
0

Hint
hint
 

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.

//题目：有n盏灯上来都是关闭的，对其进行开关操作，当进行第n次操作时，只对序号是n的倍数的灯进行开关（开→关/关→开）
//输入：灯的序号n
//输出：无穷次操作后第n盏灯的状态（0代表关，1代表开）
//思路：①当操作n次后，以后后的操作就不对第n盏灯有效了
  //    ②每当n是当前操作次数的倍数时，计数of加1（计数of每次都要初始化）
    //  ③结果为总计数除以2的余数

#include<iostream>
using namespace std;
int main()
{
    int n,i,of;
    while(cin>>n)
    {
        of=0;                //初始化
        for(i=1;i<=n;i++)
            if(n%i==0)       //步骤②
                of++;
        cout<<of%2<<endl;    //步骤③
    }
    return 0;
}