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

3
4
0

Sample Output

0
2
#include<cstdio>
typedef long long LL;
#define M  1000000007
int 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;
}