HDU 2053 Switch Game

Posted 2021-03-12 ziggystardust-pop

Switch Game

原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=2053
Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
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.

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Output

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

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Sample Input

1
5

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Sample Output

1
0

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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.
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题目大意：有n个灯泡排成一列，所有灯泡的初始状态都是关闭的，现在对这些灯泡做开关操作：在第n次操作时，号码是n的倍数的灯状态改变（开变为闭，闭变开）

 

 

尝试一下模拟第n次灯泡的状态

c语言AC代码：

#include<stdio.h>
int main()
{
        int n;
        while(scanf("%d",&n)!=EOF)
        {
            int t=0;       //初始状态为0
            for(int i=1;i<=n;i++)  //用for循环模拟第n次的灯泡状态
            {
                if(n%i==0)    
                {　　　　　　//如果n是操作次数i的倍数，则改变灯泡状态
                    t=!t; 
                }
            }
            printf("%d
",t);//输出第n个灯泡的状态
        }

        return 0;

}