hdu 3501 Calculation 2 (欧拉函数)

Calculation 2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1923    Accepted Submission(s): 812

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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//15MS    208K    517 B    G++
/*

        一开始用的是求其模数加上模数倍数，后来发现会出现重复加，
    即要用到容斥原理，不过没试过就放弃了。
        后来查了发现 n*euler(n)/2 即n及n的欧拉数积除以二的结果为
    小于n且与n互质的数的和，然后解决。 

*/
#include<stdio.h>
#define N 1000000007
int euler(int n) //直接求法 
{
    int ret=1;
    for(int i=2;i*i<=n;i++){
        if(n%i==0){
            n/=i,ret*=i-1;
            while(n%i==0){
                n/=i,ret*=i;
            }
        }
    }
    if(n>1) ret*=n-1;
    return ret;
} 
__int64 cul(__int64 n)
{
    return (n*(n+1)/2-n*euler((int)n)/2-n)%N;
}
int main(void)
{
    __int64 n;
    while(scanf("%I64d",&n),n)
    {
        printf("%I64d
",cul(n));
    }
    return 0; 
}