ACM水题堆（一）T-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 ).InputEach test case contains only a number n ( 0< n<= 10^5) in a line.OutputOutput the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).Sample Input

1
5

Sample Output

1
0

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.

这个题目主要是在题目意思的理解，输入一个数i，执行到第I个步骤，每i个步骤的将十五个1中的第I和I的倍数位的数变为上一步操作相反的状态，再输出第I个状态的数字，找规律就可以了

#include<iostream>#include<cmath>using namespace std;int main(){ int n; while (scanf("%d", &n) != EOF) {  //int con[16] = { 1 };  int i=sqrt(n);  if (pow(i, 2) == n)  {   cout << 1 << endl;  }  else  {   cout << 0 << endl;  } } return 0;}