思路:记一个公式。
给出一个N,求[1,n]中与N互质的数的和就是这个公式:
n*phi[n]/2
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<set>
#include<map>
#include<cmath>
#define INF 0x3f3f3f3f
#define ll long long
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define DEBUG cout<<endl<<"DEBUG"<<endl;
#define Max 100000
#define mod 1000000007
using namespace std;
typedef pair<ll, ll> p;
ll euler(ll n) {
ll res = n, a = n;
for(ll i = 2; i * i <= a; i++) {
if(a % i == 0) {
res = res / i * (i - 1);
while(a % i == 0)
a /= i;
}
}
if(a > 1)
res = res / a * (a - 1);
return res;
}
int main() {
ll n;
while(cin >> n && n) {
cout << ((n * (n - 1)) / 2 - (n * euler(n)) / 2) % mod << endl;
}
return 0;
}