hdoj-【3501 Calculation 2】

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3791    Accepted Submission(s): 1587

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

0
2
#include<cstdio>typedef long long LL;#define M  1000000007int main(){LL n;while(~scanf("%lld",&n),n){LL i,ans=n,temp=n,sum1,sum2;for(i=2;i*i<=n;++i){if(n%i==0)ans=ans*(i-1)/i;while(n%i==0)n/=i; } if(n!=1)ans=ans*(n-1)/n;sum1=ans*temp/2;//计算所有和n互质的数的和 sum2=temp*(temp-1)/2;//求出1-n-1的和 //printf("%lld\n",sum1);//printf("%lld\n",sum2); printf("%lld\n",(sum2-sum1)%M); } return 0;}