HDU 3501 Calculation 2

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#欧拉函数

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本文介绍HDU3501 Calculation2题目，该题要求计算小于N且与N不互质的正整数之和。通过使用欧拉公式推导出解决方案，并给出具体实现代码。

HDU 3501 Calculation 2

题目链接

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.

Ouput

For each test case, you should print the sum module 1000000007 in a line.

Sample Input
```
3
4
0
```
Sample Output
```
0
2
```

题解

    欧拉公式的引伸：小于或等于n的数中，与n互质的数的总和为：φ(x) * x / 2。(n>1)
所以：res = n*(n-1)/2-n - phi[x]*x/2

代码如下：

#include <iostream>

using namespace std;
typedef long long ll;
const int mod=1000000007;

int euler1(int n)
{
    int rea=n,sum=0;
    for(int i=2; i*i<=n; i++)
        if(n%i==0)
        {
            rea=rea-rea/i;
            do
                n/=i;
            while(n%i==0);
        }
    if(n>1)
        rea=rea-rea/n;
    return rea;
}
int main()
{
    ll n;
    while(cin>>n,n){
        ll s=(ll(n*(n-1)/2)%mod-ll(n*euler1(n)/2)%mod+mod)%mod;
        cout<<s<<endl;
    }
    return 0;
}