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

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 1000000007long 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;}