Codeforces 1499D - The Number of Pairs （数学）

Educational Codeforces Round 106 (Rated for Div. 2) D. The Number of Pairs

Title

Given three positive integers $c,d,x$ , ask how many pairs of positive integers$(a,b)$ , satisfying
$$c\cdot lcm-d\cdot gcd=x$$

limit

$1\le T\le 10^4$

$1\le c,d,x\le 10^7$

Ideas

For the original formula
$$c\cdot lcm-d\cdot gcd=x$$
by$gcd$ and$LCm$ mathematical properties, it is easy to obtain a conclusion:$lcm$ must begcd gcdThe multiple of$g$$c$$d$ .

Then the above formula can be transformed into
$$lcm=\frac{x+d\cdot gcd}{c}$$
The right side must be an integer, that is $x+d\cdot gcd$ must be$c$ multiple

Then we can $O\left(\sqrt{x}\right)$ Enumxxall factors of$x$ as$gcd$

Since $c,d,x\; is$ known, if the above conditions are met,$gcd,c,d,x$ directly calculate the matched$lcm$

The problem has now been transformed into "After the known $gcd$ andlcm lcmUnder the premise of$l$$c$$m$ , find how many pairs of integers satisfy the condition"

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From a mathematical point of view

$gcd(a,b)$ is$a,$the intersection of the prime factors of$b$

$lcm(a,b)$ is$a,the$ union of the prime factors of$b$

Then $\frac{lcm}{gcd}$Is the product of the extra prime factors

For $\frac{lcm}{gcd}$Of each quality factor, regardless of the number , should be fully positioned $\frac{a}{gcd}$Or all of them are located in $\frac{b}{gcd}$in

Otherwise, these prime factors are in $a$ and$If\; b$ appears at the same time, it will be counted as$In\; gcd$ , the situation does not match at this time

So if we want to construct $a,b$ Two numbers, each prime factor can only pick one of them and multiply it in

设$\frac{lcm}{gcd}$The number of prime factors is $n$ , the integer pairs that can be constructed have a total of$2^n$ kinds

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Practice can be obtained, the online prime factorization algorithm will be perfect TLE 15

So consider offline processing of the number of prime factor types that each number has

The writing method here is similar to the Ehrlich prime number sieve method, see init init in the code section for details$init$ function

注意$\frac{\frac{x}{gcd}+d}{c}$In theory, the maximum possible data range is $2\cdot 10^7$

Code

(1075ms/2000ms)

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

const int MAXN=2e7;
int r[MAXN+50]; //记录每个数字的质因子种类数

void init()
{
    
    
    for(int i=2;i<=MAXN;i++)
        if(!r[i]) //判断i是否为素数
        {
    
    
            for(int j=i;j<=MAXN;j+=i)
                r[j]++; //i在范围内的倍数的质因子种类数++
        }
}

ll c,d,x,ans;

void add(int gcd)
{
    
    
    ll lcm=x+d*gcd;
    if(lcm%c==0)
    {
    
    
        lcm/=c;
        if(lcm%gcd==0)
            ans+=1LL<<r[lcm/gcd];
    }
}

void solve()
{
    
    
    ans=0;
    cin>>c>>d>>x;
    int s=sqrt(x);
    for(int i=1;i<=s;i++)
    {
    
    
        if(x%i==0) //枚举x的因子i
        {
    
    
            add(i);
            if(i*i!=x)
                add(x/i);
        }
    }
    cout<<ans<<'\n';
}

int main()
{
    
    
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    init();
    int T;cin>>T;while(T--)
        solve();
    return 0;
}