hdu 3501 Calculation 2

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Calculation 2Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2135    Accepted Submission(s): 898

Problem DescriptionGiven 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. 
InputFor 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. 
OutputFor each test case, you should print the sum module 1000000007 in a line. 
Sample Input

340

 
Sample Output

02

 

#include <iostream>#include <cstdio>using namespace std;#define LL __int64LL eular(LL n)   // 求歐拉函數{    LL i,res=n;    for(i=2;i*i<=n;i++){if(n%i==0){res=res/i*(i-1);while(n%i==0)n/=i;}}if(n>1)res=res/n*(n-1);return  res;}int main(){    LL n;    while(scanf("%I64d",&n)&&n!=0)    {        if(n==1)        {            printf("0\n");            continue;        }        LL k=eular(n);//printf("%I64d\n",k);     //  n的歐拉函數值//printf("%I64d\n",n*k/2);  // 小於n 且與n互質的數 之和        printf("%I64d\n",((n-1)*n/2-n*k/2)%1000000007);  //小於n 且與n不互質的數 之和     }    return 0;}