原题链接
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：
给一个整数n,求小于n且与n不互质的数的和
欧拉函数φ(n)的值为小于n且与n互质的数的个数（见文末），再想到gcd的一个性质如果gcd(n,i)=1,则gcd(n,n-i)=1,可以看出与n互质的数成对出现的，即欧拉函数的值为偶数(但φ(1)=1),且每一对的和都为n。
这样解题思路就出来了，用1-n的和减去n*φ(n)/2,即为答案
#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 ",ans ); return 0; }
欧拉函数φ：指从1到n-1与n互质的数的个数，特别的定义φ(1)=1下面给出其公式
其中pi为n的素因子。