hdu 3501 Calculation 2

Posted 2020-10-01

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4551    Accepted Submission(s): 1879

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

 

 

Recommend

 

题意：给定N 求小于N 且与N不互素的数的和

欧拉函数性质 ： 与x互素的数的和=x*φ(x)/2

屠龙宝刀点击就送

#include <ctype.h>
#include <cstdio>
void read(long long &x)
{
    x=0;bool f=0;register char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch==‘-‘) f=1;
    for(; isdigit(ch);ch=getchar()) x=(x<<3)+(x<<1)+ch-‘0‘;
    x=f?(~x)+1:x;
}
long long a=1,k;
long long getphi(long long n)
{
    long long ans=n;
    if(n%2==0)
    {
        while(n%2==0) n/=2;
        ans/=2;
    }
    for(int i=3;i*i<=n;i+=2)
    {
        if(n%i==0)
        {
            while(n%i==0) n/=i;
            ans=ans/i*(i-1);
        }
    }
    if(n>1) ans=ans/n*(n-1);
    return ans;
}
int main()
{
    for(;a;)
    {
        read(k);
        if(!k) break;
        long long sum=(long long)k*(k-1)/2;sum-=(long long)getphi(k)*k/2;
        printf("%lld\n",sum% 1000000007);//再忘了取模就剁手！！
    }
    return 0;
}