Hduoj2053 【水题】

Switch Game

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10929    Accepted Submission(s): 6661

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

15

 

Sample Output

10
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.

 

Author

LL

 

Source

校庆杯Warm Up 

#include<stdio.h>int main(){int i, j, k;while(scanf("%d", &k) != EOF){j = 0;if(k == 1){printf("1\n");continue;}for(i = 2; i < k; i++){if(k % i == 0){if(j == 0)j = 1;elsej = 0;}}printf("%d\n", j); } return 0;} 

//最简代码

#include<cstdio>#include<cmath>int main(){int n,k;while(~scanf("%d",&n)){k=sqrt(n);if(k*k==n)printf("1\n");else printf("0\n");}return 0;}