Original title link
http://acm.hdu.edu.cn/showproblem.php?pid=3501
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
题意分析：
To an integer number n, and n is less than the demand and n is not prime, and
Euler function [Phi] (n) value is less than n and prime with n the number of counts (end of the text), then think of a nature, if gcd gcd (n, i) = 1, then the gcd (n, Ni) = 1, we can see the quality and number of paired cross-n, i.e., the value is even Euler function ([Phi], but (1) = 1), and each pair and are to n.
Such problem-solving ideas came out with a 1-n and subtract n * φ (n) / 2, is the answer
#include <iostream> #include <stdio.h> #include <algorithm> #include <string.h> #include <vector> #include <math.h> #include <map> #include <queue> #include <set> using namespace std; typedef long long ll; const int maxn=1e5; const int mod=1e9+7; ll a[maxn]; //o(sqrt(n))求欧拉函数的值 int Phi(int n){ int m=(int)sqrt(n+0.5); int ans=n; for(int i=2;i<=m;i++){ if(n%i==0){ ans=ans/i*(i-1); while(n%i==0) n/=i; } } if(n>1) ans=ans/n*(n-1); return ans; } int main() { ll n; while(scanf("%lld",&n)!=-1&&n) { ll ans=(n*(n-1)/2)%mod; ans-=(n*Phi(n)/2)%mod; cout<<(ans+mod)%mod<<endl; } //printf("%lld\n",ans ); return 0; }
Euler function φ: refers to the number of prime numbers, in particular the definition of φ (1) = 1 from Equation 1 given below it to n-1 and n
Wherein pi is the prime factors of n.
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