欧拉函数：HDU3501-Calculation 2

Calculation 2

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

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

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解题心得：

1. 题意是叫你在1-n之间找出所有和n互质的数，然和求和。就是一个简单的欧拉函数的拓展，n的数中，与n互质的数的总和为：φ(x) * x / 2  (n>1)。其实就是n*（互质的数的个数）/2。知道了这些就很简单了。

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 1e9+7;

//先将所有的互质的数的个数给求出来
ll eular(ll n)
{
    ll ans = n;
    for(int i=2;i<=sqrt(n);i++)
    {
        if(n%i == 0)
        {
            ans = ans/i*(i-1);
            ans %= mod;

            while(n%i == 0)
            {
                n /= i;
            }
        }
    }
    if(n > 1)
        ans = ans/n*(n-1);
    return ans;
}

int main()
{
    ll n;
    while(scanf("%lld",&n) && n)
    {
        ll sum = (n-1)*n/2;
        ll sum2 = n*eular(n)/2;//按照这个公式可以知道所有的互质的数的和
        printf("%lld
",(sum-sum2)%mod);
    }
    return 0;
}