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
phi(n)表示比n小且与n互素的整数的个数，而phi(n)*n/2表示比n小且与n互素的整数的和。
因此算一下前n个数的和，然后用容斥原理就行了。
#include<iostream> #include<cmath> using namespace std; typedef long long LL; #define mod 1000000007 LL euler_phi(LL n) //欧拉函数 { LL m=sqrt(n+0.5),ans=n,i=2; for(;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(cin>>n&&n) { LL ans=n*(n+1)/2-n-euler_phi(n)*n/2; //容斥原理 cout<<ans%mod<<endl; } return 0; }