题解报告：hdu 3501 Calculation 2 （欧拉函数的扩展）

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
解题思路：求小于n且与n不互质的所有数之和，公式：n*(n-1)/2-n*Euler(n)/2。
AC代码：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <map>
 #include <vector>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int maxn = 1e6+;
 const LL mod = ;
 LL n;
 LL get_Euler(LL x){
     LL res = x; ///初始值
     for(LL i = 2LL; i * i <= x; ++i) {
         if(x % i == ) {
             res = res / i * (i - ); ///先除后乘，避免数据过大
             while(x % i == ) x /= i;
         }
     }
     if(x > 1LL) res = res / x * (x - ); ///若x大于1，则剩下的x必为素因子
     return res;
 }

 int main(){
     while(cin >> n && n) {
         cout << (n * (n - ) /  - n * get_Euler(n) / ) % mod << endl;
     }
     return ;
 }