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HDOJ 4983 Goffi and GCD

2024-07-19 09:30:51   224人阅读


题意：
给你 N 和 K,问有多少个数对满足 gcd(N-A, N) * gcd(N - B, N) = N^K

分析：
由于 gcd(a, N) <= N,于是 K>2 都是无解,K=2 只有一个解 A=B=N,只要考虑 K = 1 的情况就好了
其实上式和这个是等价的 gcd(A, N) * gcd(B, N) = N^K,我们枚举 gcd(A, N) = g,那么gcd(B, N) = N / g。问题转化为统计满足 gcd(A, N) = g 的 A 的个数。这个答案就是 ?(N/g)
只要枚举 N 的 约数就可以了。答案是 Σ?(N/g)*?(g) g | N
计算 ? 可以递归,也可以直接暴力计算,两个都可以。


Goffi and GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 361    Accepted Submission(s): 118


Problem Description
Goffi is doing his math homework and he finds an equality on his text book: \(\gcd(n - a, n) \times \gcd(n - b, n) = n^k\).

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

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 \(k\) (\(1 \le n, k \le 10^9\)).
 

Output
For each test case, output a single integer indicating the number of (\(a, b\)) modulo \(10^9+7\).
 

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
 



#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>

using namespace std;

const int mod=(1e9+7);

long long int  euler_phi(int n)
{
    int m=(int)sqrt(n+0.5);
    long long int ans=n;
    for(int i=2;i<=m;i++)
    if(n%i==0)
    {
        ans=ans/i*(i-1);
        while(n%i==0)
            n/=i;
    }
    if(n>1)
        ans=ans/n*(n-1);
    return ans;
}

int n,k;

int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(n==1) puts("1");
        else if(k>2) puts("0");
        else if(k==2) puts("1");
        else
        {
            long long int ans=0;
            for(int g=1;g<=n;g++)
            {
                if(n<g*g) break;
                if(n%g==0)
                {
                    if(g!=n/g)
                        ans=(ans+(euler_phi(n/g)*euler_phi(g))%mod*2)%mod;
                    else
                        ans=(ans+(euler_phi(n/g)*euler_phi(g))%mod)%mod;
                }
            }
            printf("%I64d\n",ans%mod);
        }
    }
    return 0;
}



HDOJ 4983 Goffi and GCD


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