HDU 2608 0 or 1

0 or 1
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3708 Accepted Submission(s): 1077

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
3
1
2
3

Sample Output
1
0
0

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

Author
yifenfei

题目意思不就是分别求下T[i],然后再对T[i]求和，求其s%2的值，这个数据量蛮大的，一遍遍过去肯定是要超时的，然后找规律啊，发现这个s竟然是i* i<= x和2* i* i<=x的个数，其实反向求也是正确的，毕竟只有0和1，我是想不出来他们怎么0ms通过的，可能打表吧，处理一下，数据量大访问就可以了，不浪费多少时间的，存完全平方数，2^31大概是2^15的平方，其实也就1024*32，我想了想处理sqrt用了二分什么的发现没有卵用，恐怕他们用的就是打表了省时间

#include <stdio.h> int main(  ){      int T;    scanf( "%d", &T );    while( T-- )    {        long long x;        int n=-1;        scanf( "%lld", &x );        for( long long i= 1; i* i<= x; ++i )        {            if( 2* i* i>x )            n=-n;        }        printf("%d\n",n==1?1:0);    }    return 0;}