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+5;
const LL mod = 1000000007;
LL n;
LL get_Euler(LL x){
    LL res = x; ///初始值
    for(LL i = 2LL; i * i <= x; ++i) {
        if(x % i == 0) {
            res = res / i * (i - 1); ///先除后乘，避免数据过大
            while(x % i == 0) x /= i;
        }
    }
    if(x > 1LL) res = res / x * (x - 1); ///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 - 1) / 2 - n * get_Euler(n) / 2) % mod << endl; 
    }
    return 0;
}