HDOJ题目2608 0 or 1（数学）

0 or 1

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2668    Accepted Submission(s): 688

Problem 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
Hint S(3) = T(1) + T(2) +T(3) = 1 + (1+2) + (1+3) = 8     S(3) % 2 = 0 
 

 

Author

yifenfei

 

Source

奋斗的年代

 

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ac代码

#include<string.h>#include<stdio.h>int main(){int t;scanf("%d",&t);while(t--){__int64 n,i,sum=0;scanf("%I64d",&n);for(i=1;i*i<=n;i++){sum++;if(2*i*i<=n)sum++;}if(sum%2==0)printf("0\n");elseprintf("1\n");}}