HDU 3501 Calculation 2------欧拉函数变形

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1750    Accepted Submission(s): 727

Problem Description

Given a positive integer N, your task is to calculate the sum of the positive integers less than N which are not coprime to N. A is said to be coprime to B if A, B share no common positive divisors except 1.

 

Input

For each test case, there is a line containing a positive integer N(1 ≤ N ≤ 1000000000). A line containing a single 0 follows the last test case.

 

Output

For each test case, you should print the sum module 1000000007 in a line.

 

Sample Input

3

4

0

Sample Output

0

2

/*
欧拉函数一个小小的性质：

求小于n并且与n互质的数的和为：n*phi[n]/2；

这个好像以前有点印象的。

*/

#include<stdio.h>


__int64 Euler(__int64 n)
{
    __int64 temp=n,i;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            while(n%i==0)
            n=n/i;
            temp=temp/i*(i-1);
        }
    }
    if(n!=1) temp=temp/n*(n-1);
    return temp;
}

int main()
{
    __int64 n,m;
    __int64 k;
    while(scanf("%I64d",&n)>0)
    {
        if(n==0)break;
        m=Euler(n);
        k=m*n/2;
        k=n*(n+1)/2-n-k;
        //刚开始，只有k为__in64，错了。
        //这里相乘之后，可能会超过int范围。倒置溢出
        k=k%1000000007;
        printf("%I64d
",k);
    }
    return 0;
}