HDOJ Calculation 2 3501（eular函数）

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2511    Accepted Submission(s): 1048

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
   
   
   
   

 

Author

GTmac

 

Source

2010 ACM-ICPC Multi-University Training Contest（7）——Host by HIT

 

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欧拉函数性质：N>1，不大于N且和N互素的所有正整数的和是 1/2*M*eular(N)。

#include<stdio.h>
#include<algorithm>
using namespace std;
typedef __int64 LL;
LL phi(LL n){
	LL i,res=n;
	for(i=2;i*i<=n;i++)
	    if(n%i==0){
	    	res=res/i*(i-1);
	    	while(n%i==0)n/=i;
		}
	if(n>1) res=res/n*(n-1);
	return res;
}
int main()
{
	LL N;
	while(scanf("%I64d",&N),N){
		LL x=(N*(N-1)/2-N*phi(N)/2)%1000000007;
		printf("%I64d\n",x);	
	}
	return 0;
}