hdoj Calculation 2 3501 （欧拉函数）

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3141    Accepted Submission(s): 1300


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, you should print the sum module 1000000007 in a line.

[align=left]Sample Input[/align]

3
4
0

[align=left]Sample Output[/align]

0
2//题意：求小于且不与n互质的数之和。用欧拉函数先求出与n互质的数之和，再用总和减去就行了。Hint： 欧拉函数求得的与n互质的值为n*ephi(n)/2;

#include<stdio.h>
#include<string.h>
#include<math.h>
#define M 1000000007
long long ephi(long long n)
{
int m=(int)sqrt(n+0.5);
long long ans=n;
for(int i=2;i<=m;i++)
if(n%i==0)
{
ans=ans/i*(i-1);
while(n%i==0)
n/=i;
}
if(n>1)
ans=ans/n*(n-1);
return ans;
}
int main()
{
long long n;
while(scanf("%lld",&n),n)
{
long long ans=((n*(n-1)/2)-((n*ephi(n)/2)))%M;
printf("%lld\n",ans);
}
return 0;
}