数论 0OR1

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I - 0 or 1
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 2608

Description

Solving problem is a interesting thing. Yifenfei like to slove different problem,because he think it is a way let him more intelligent. But as we know,yifenfei is weak in math. When he come up against a difficult math problem, he always try to get a hand. Now the problem is coming! Let we
define T(n) as the sum of all numbers which are positive integers can divied n. and S(n) = T(1) + T(2) + T(3)…..+T(n). 
 

Input

The first line of the input contains an integer T which means the number of test cases. Then T lines follow, each line consists of only one positive integers n. You may assume the integer will not exceed 2^31.
 

Output

For each test case, you should output one lines of one integer S(n) %2. So you may see the answer is always 0 or 1 .
 

Sample Input

3123
 

Sample Output

100

Hint

S(3) = T(1) + T(2) +T(3) = 1 + (1+2) + (1+3) = 8
     S(3) % 2 = 0  
 
#include<stdio.h>#include<math.h>int main(){int n,m,s;scanf("%d",&n);while(n--){scanf("%d",&m);s=(int) sqrt(m*1.0)+ (int )sqrt(m/2.0);printf("%d\n",s%2);}}


0 0