HDU 3501 alculation 2 【欧拉函数】

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2379    Accepted Submission(s): 1003

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

340

 

Sample Output

02

/*
题目大意：求所有小于N且与N为非互质数的和


小于N且与N互质的数的和 = 小于N的数的和 - 小于N且与N为互质数的和 


求所有小于N且与N为互质数的和 
1.欧拉函数可求与N互质的数的个数
2.若已知m与n互质，则n-m也与n互质 
*/
欧拉函数证明： 
http://hi.baidu.com/ldante/item/28042208e1f139133b53eefc 

#include<cstdio>#define mod 1000000007__int64 eular(__int64 n)//欧拉函数模板 {    __int64 ret=1;    for(__int64 i=2; i*i<=n; i++)    {        if(n%i==0)        {            n/=i; ret*=i-1;            while(n%i==0)            {                n/=i; ret*=i;            }        }    }    if(n>1) return ret*=n-1;    return ret;}int main(){    __int64 ans,n;    while(scanf("%I64d",&n)&&n)    {        ans=n*(n-1)/2;        printf("%I64d\n",(ans-n*eular(n)/2)%mod);    }    return 0;}