2020牛客暑期多校训练营（第三场）F - Fraction Construction Problem

Description

给定 \(a,b \le 2\times 10^6\)，构造 \(c,d,e,f\) 满足 \(c/d-e/f=a/b\)，且 \(d,f<b\)，\(c,d,e,f \le 4 \times 10^{12}\)

Solution

原式变形得

\[\frac {cf-de} {df} = \frac a b =\frac {a'} {b'},\quad a'=\frac a {(a,b)}, \quad b'=\frac b {(a,b)} \]

若 \(b \neq b'\)，则令 \(c=a'+b', d=b', e=1,f=1\) 即可

若 \(b=b'\)，考虑将 \(b\) 分解为两个互质的部分 \(d,f\) 且 \(d\neq b, f\neq b\)

以 \(c,e\) 为未知量，解不定方程 \(cf-ed=a\) 即可

为了方便，先将 \(c,(-e)\) 解出，然后对于 \(c,e\)，当有负数时，循环将 \(c+=d,e+=f\) 即可

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

#define int long long
#define LL long long

vector <int> prime;

LL gcd(LL a,LL b){
    if(b==0) return a;
    else return gcd(b,a%b);
}

LL exgcd(LL a,LL b,LL &x,LL &y){
    if(b==0){
        x=1;y=0;
        return a;
    }
    LL r=exgcd(b,a%b,x,y);
    LL t=x;
    x=y;
    y=t-a/b*y;
    return r;
}

void solve()
{
    int a,b;
    scanf("%lld%lld",&a,&b);
    if(__gcd(a,b)>1)
    {
        int g=__gcd(a,b);
        a/=g; b/=g;
        printf("%lld %lld 1 1\n",a+b,b);
    }
    else
    {
        for(int d:prime) if(d*d<=b && b%d==0)
        {
            int f=b/d;
            int d0=d;
            while(f%d0==0) f/=d0, d*=d0;
            if(f==1) continue;
            int c,e;
            exgcd(f,d,c,e);
            e=-e;
            while(c+d*10000<=0 || e+f*10000<=0)
            {
                c+=d*10000;
                e+=f*10000;
            }
            while(c+d*1000<=0 || e+f*1000<=0)
            {
                c+=d*1000;
                e+=f*1000;
            }
            while(c+d*100<=0 || e+f*100<=0)
            {
                c+=d*100;
                e+=f*100;
            }
            while(c+d*10<=0 || e+f*10<=0)
            {
                c+=d*10;
                e+=f*10;
            }
            while(e<=0 || c<=0)
            {
                c+=d;
                e+=f;
            }
            c*=a;
            e*=a;
            if((c*f-d*e)*b!=a*d*f) cout<<"failed"<<endl;
            printf("%lld %lld %lld %lld\n",c,d,e,f);
            return;
        }
        /*for(int f=2;f*f<=b;f++) if(b%f==0)
        {
            int d=b/f;
            if(d==1) continue;
            if(__gcd(f,d)>1) continue;
            int c,e;
            exgcd(f,d,c,e);
            e=-e;
            while(e<=0 || c<=0)
            {
                c+=d;
                e+=f;
            }
            c*=a;
            e*=a;
            if((c*f-d*e)*b!=a*d*f) cout<<"failed"<<endl;
            cout<<c<<" "<<d<<" "<<e<<" "<<f<<endl;
            return;
        }*/
        puts("-1 -1 -1 -1");
    }
}

signed main()
{
    ios::sync_with_stdio(false);

    for(int i=2;i<=1500;i++)
    {
        int flag=1;
        for(int j=2;j<i;j++)
        {
            if(i%j==0)
            {
                flag=0;
                break;
            }
        }
        if(flag) prime.push_back(i);
    }

    int t;
    scanf("%lld",&t);
    while(t--)
    {
        solve();
    }
}