hdu5528 Count a * b

Count a * b

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 262    Accepted Submission(s): 132

Problem Description

Marry likes to count the number of ways to choose two non-negative integers  a and  b less than  m to make  a×b mod  m≠0.

Let's denote  f(m) as the number of ways to choose two non-negative integers  a and  b less than  m to make  a×b mod  m≠0.

She has calculated a lot of  f(m) for different  m, and now she is interested in another function  g(n)=∑m|nf(m). For example,  g(6)=f(1)+f(2)+f(3)+f(6)=0+1+4+21=26. She needs you to double check the answer.



Give you  n. Your task is to find  g(n) modulo  264.

 

Input

The first line contains an integer  T indicating the total number of test cases. Each test case is a line with a positive integer  n.

1≤T≤20000
1≤n≤109

 

Output

For each test case, print one integer  s, representing  g(n) modulo  264.

 

Sample Input

2 6 514

 

Sample Output

26 328194

 

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

 

#include
const int maxn=100001;
using namespace std;
bool bb[maxn];
int prim[maxn],tot,v[maxn],s[maxn],t[maxn],n;
void prepare(){
    bb[1]=1;v[1]=1;t[1]=1;
    for(int i=2;i<=maxn;i++){
        if(!bb[i]){
            prim[++tot]=i;
            v[i]=2*i-1;
            s[i]=1;
            t[i]=i;
        }
        for(int j=1;j<=tot&&prim[j]*i1){
        ans1*=(1LLU*n*n+1);
        ans2*=2*n;
    }
   //cout<