Hdu--4983（数论，欧拉函数）

2014-09-04 14:10:38

Goffi and GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 584    Accepted Submission(s): 208

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

 

思路：搬运一下hdu题解：

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 ɸ(N/g)
只要枚举 N 的 约数就可以了。答案是 Σɸ(N/g)*ɸ(g) g | N
计算 ɸ 可以递归,也可以直接暴力计算,两个都可以。

 

/*************************************************************************
    > File Name: 1003.cpp
    > Author: Nature
    > Mail: 564374850@qq.com
    > Created Time: Thu 04 Sep 2014 11:40:55 AM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int mod = (1e9) + 7;

ll n,k;
ll ans;

ll Phi(ll v){
    ll tmp = v;
    ll m = sqrt(v + 0.5);
    for(ll i = 2; i <= m; ++i) if(v % i == 0){
        tmp = tmp / i * (i - 1);
        while(v % i == 0) v /= i;
        if(v == 1) break;
    }
    if(v != 1) tmp = tmp / v * (v - 1);
    return tmp;
}

int main(){
    while(scanf("%I64d%I64d",&n,&k) != EOF){
        if(n == 1) printf("1\n");
        else if(k > 2) printf("0\n");
        else if(k == 2) printf("1\n");
        else{
            ans = 0;
            ll m = sqrt(n + 0.5);
            for(ll i = 1; i <= m; ++i) if(n % i == 0){
                ll sum = Phi(i) * Phi(n / i) * 2;
                ans = (ans + sum) % mod;
            }
            if(m * m == n){
                ll sum = Phi(m);
                ans = (ans - sum * sum + mod) % mod;
            }
            printf("%I64d\n",ans);
        }
    }
    return 0;
}