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
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61 5
Sample Output
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311 0Consider 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.Hinthint
Author
LL
C语言AC代码
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16#include<stdio.h> #include<math.h> int main() { int n; double t; while(scanf("%d",&n)!=EOF) { t=sqrt(n*1.0); if(t==(int)t) printf("1n"); else printf("0n"); } return 0; }
思路:1 模拟所有情况,超时
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n的
的状态由n的约数个数决定。奇数时,开启,偶数时,关闭。
3 约数的个数为奇数时,n为平方数(平方数的两个约数相同),由此,判断n是否为平 方数即可。
最后
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