P1072 Hankson的趣味题

Solution ((90\%))

注意到(operatorname{lcm}(b_0,x) = b_1)，所以(x mid b_1)，枚举(b_1)的所有约数并暴力计算。
时间复杂度(mathcal{O}(n sqrt{b_1} log{b_1}))

Solution ((100\%))

枚举质数(p mid d_1)，令(m_1)为(a_0)中包含(p)因子的个数，(m_2,m_3,m_4)同理。
进行分类讨论。

# include <bits/stdc++.h>
using namespace std;

const int N = 1e6 + 5;

int n;

int ans = 1;
int a0,a1,b0,b1;

template <typename T> void read(T &x)
{
    int w = 1;
    x = 0;
    char ch = getchar();
    while(!isdigit(ch))
    {
        if(ch == '-') w = -1;
        ch = getchar();
    }
    while(isdigit(ch))
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    x *= w;
    return;
}

void solve(int p)
{
    int m1 = 0,m2 = 0,m3 = 0,m4 = 0;
    while(a0 % p == 0)
    {
        a0 /= p;
        ++m1;
    }
    while(a1 % p == 0)
    {
        a1 /= p;
        ++m2;
    }
    while(b0 % p == 0)
    {
        b0 /= p;
        ++m3;
    }
    while(b1 % p == 0)
    {
        b1 /= p;
        ++m4;
    }
    // printf("p = %d,m1 = %d,m2 = %d,m3 = %d,m4 = %d
",p,m1,m2,m3,m4);
    if(m1 > m2 && m3 < m4)
    {
        if(m2 != m4) ans = 0;
    }
    else if(m1 > m2 && m3 == m4) 
    {
        if(m2 > m4) ans = 0;
    }
    else if(m1 == m2 && m3 < m4) 
    {
        if(m2 > m4) ans = 0;
    }
    else if(m1 == m2 && m3 == m4)
    {
        if(m2 > m4) ans = 0;
        else ans *= (m4 - m2 + 1);
    }
    else ans = 0;
    return;
}
int v[N],prime[N],tot;

void Get_prime(int n)
{
    for(int i = 2; i <= n; i++)
    {
        if(v[i] == 0)
        {
            v[i] = i;
            prime[++tot] = i;
        }
        for(int j = 1; j <= tot && prime[j] * i <= n; j++)
        {
            if(prime[j] > v[i]) break;
            v[prime[j] * i] = prime[j];
        }
    }
    return;
}

int main(void)
{
    read(n);
    Get_prime(50000);
    int _b1 = b1;
    while(n--)
    {
        ans = 1;
        read(a0),read(a1),read(b0),read(b1);
        for(int i = 1; i <= tot; i++)
        {
            if(_b1 % prime[i] == 0) solve(prime[i]);
            else break;
            if(ans == 0) break;
        }
        if(b1 != 1) solve(b1);
        printf("%d
",ans);
    }
    return 0;
}