（跳过了 怕了）Hankson的趣味题

题目 Hankson的趣味题

思路分析

代码实现

#include <iostream>

#include <algorithm>

#include <vector>

using namespace std;

typedef long long LL;

typedef pair<int, int> PII;

const int N = 45000, M = 50;

int primes[N], cnt;

bool st[N];

PII factor[M];

int cntf;

int divider[N], cntd;

void get_primes(int n)

{

    for (int i = 2; i <= n; i ++ )

    {

        if (!st[i]) primes[cnt ++ ] = i;

        for (int j = 0; primes[j] <= n / i; j ++ )

        {

            st[primes[j] * i] = true;

            if (i % primes[j] == 0) break;

        }

    }

}

int gcd(int a, int b)

{

    return b ? gcd(b, a % b) : a;

}

void dfs(int u, int p)

{

    if (u > cntf)

    {

        divider[cntd ++ ] = p;

        return;

    }

    for (int i = 0; i <= factor[u].second; i ++ )

    {

        dfs(u + 1, p);

        p *= factor[u].first;

    }

}

int main()

{

    get_primes(N);

    int n;

    scanf("%d", &n);

    while (n -- )

    {

        int a0, a1, b0, b1;

        scanf("%d%d%d%d", &a0, &a1, &b0, &b1);

        int d = b1;

        cntf = 0;

        for (int i = 0; primes[i] <= d / primes[i]; i ++ )

        {

            int p = primes[i];

            if (d % p == 0)

            {

                int s = 0;

                while (d % p == 0) s ++, d /= p;

                factor[ ++ cntf] = {p, s};

            }

        }

        if (d > 1) factor[ ++ cntf] = {d, 1};

        cntd = 0;

        dfs(1, 1);

        int res = 0;

        for (int i = 0; i < cntd; i ++ )

        {

            int x = divider[i];

            if (gcd(x, a0) == a1 && (LL)x * b0 / gcd(x, b0) == b1)

            {

                res ++ ;

            }

        }

        printf("%d\n", res);

    }

    return 0;

}

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