题目:
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
欧拉函数模板:
先求出φn;
由gcd(n,i)=1得gcd(n,n-i)=1,则与n互素的数之和=φ(n)*n/2;
则ans=小于n的数之和-φ(n)*n/2(注意这里不能为“-φ(n)/2*n”或“-n/2*φ(n)”,先/2可能变成0)。
以下为代码:
#include<iostream> #include<cstdio> #include<cmath> using namespace std; long long n,res,ans; int main() { while(1) { scanf("%lld",&n); if(!n)return 0; res=n; int nn=n; for(int i=2;i*i<=nn;i++) { if(nn%i==0) { res-=res/i;//减去其中素因子i的倍数的个数 // res=res/i*(i-1); while(nn%i==0)nn/=i;//提走素因子i } } if(nn>1)res-=res/nn;//不能是res--! // if(nn>1)res=res/n*(n-1); ans=(n*(n-1)/2-n*res/2)%1000000007; printf("%lld\n",ans); } return 0; }