标签：cannot follow 模拟 c代码 condition 语言 put who hang
Switch Game
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\n",t);//输出第n个灯泡的状态
}
return 0;
}
HDU 2053 Switch Game
标签：cannot follow 模拟 c代码 condition 语言 put who hang
原文地址：https://www.cnblogs.com/ziggystardust-pop/p/12259508.html
