hdu 4983 Goffi and GCD（欧拉函数）

Problem Description

Goffi is doing his math homework and he finds an equality on his text book: gcd(n−a,n)×gcd(n−b,n)=nk.

Goffi wants to know the number of (a,b) satisfy the equality, if n and k are given and 1≤a,b≤n.

Note: gcd(a,b) means greatest common divisor of a and b.

 

Input

Input contains multiple test cases (less than 100). For each test case, there's one line containing two integers n and k (1≤n,k≤109).

 

Output

For each test case, output a single integer indicating the number of (a,b) modulo 109+7.

 

Sample Input

2 1 
3 2

 

Sample Output

2
1

Hint

For the first case, (2, 1) and (1, 2) satisfy the equality.

 

Source

BestCoder Round #6

 发现自己欧拉函数都给忘记了，所有赶紧补题。。。

1、k!=1时情况很简单，记住将if(k==2 || n==1)这个特判放在if(k>2)的前面，因为这个WA了很久，各种原因自己思考。

2、下面讨论k=1时情况。x=gcd(n-a,n),则n/x=gcd(n-b,n),因为n-a可以取到0...n-1也就是1....n，所以完全可以去掉n-这个限制条件，即gcd(a,n)=x、gcd(b,n)=n/x时个数，因为a<=n，所以gcd(a,n)的个数=u[n/x],u是欧拉函数。所以原式等于sigma(u[n/x]*u[x])其中x是n的约数。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MOD 1000000007
#define ll long long
ll eular(ll n)
{
    ll res=1;
    for(ll i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            n/=i,res*=i-1;
            while(n%i==0)
            {
                n/=i;
                res*=i;
            }
        }
    }
    if(n>1) res*=n-1;
    return res;
}
ll n,k;
int main()
{
    while(scanf("%I64d%I64d",&n,&k)==2)
    {
        if(k==2 || n==1)
        {
            printf("1
");
            continue;
        }
        if(k>2)
        {
            printf("0
");
            continue;
        }
        
        ll ans=0;
        for(ll i=1;i*i<=n;i++)
        {
            if(n%i==0)
            {
                if(i*i!=n)
                  ans=(ans+eular(n/i)*eular(i)*2)%MOD;
                else 
                    ans=(ans+eular(n/i)*eular(i))%MOD;
            }
        }
        printf("%I64d
",ans);

    }
    return 0;
}