HDU_2053
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
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.
Hint
hint1 #include <cstdio> 2 int main() 3 { 4 int n, ans; 5 while(~scanf("%d",&n)) 6 { 7 ans=1; 8 for(int i=1;i<=n;i++) 9 { 10 if(n%i==0) 11 { 12 ans=-1*ans; 13 } 14 } 15 printf(ans==1?"0\n":"1\n"); 16 } 17 return 0; 18 }
——现在的努力是为了小时候吹过的牛B!!