Hankson's Interesting questions

Dr. Hanks is a well-known expert in the field of BT (Bio-tech, Biotechnology) whose son is named Hankson. Now, Hankson, who just came home from school, is thinking about an interesting question.

Today in class, the teacher explained how to ask for two positive integers c1 and C2 greatest common divisor and least common multiple. Now that Hankson thinks he has mastered this knowledge, he begins to think of a "inverse problem" of a "number of conventions" and "common multiple", the question of which is: a known positive integer a0,a1,b0,b1, an unknown positive integer x satisfies:

The greatest common divisor of 1.x and A0 is A1;
The least common multiple of 2.x and B0 is B1.
The "Inverse problem" of Hankson is to find the positive integer x that satisfies the condition. But after a little deliberation, he found that such an x is not unique and may not even exist. So instead he began to think about how to solve the number of x that satisfies the condition. Please help him to solve the problem by programming.

First, turn the conditions into mathematical formulas.

A1=GCD (X,A0);

B1=LCM (X,B0);

Obviously you'll find GCD is better than LCM, so let's convert the second one.

First you need to know a theorem a*b=gcd (a, B) *LCM (A, b);

So the second Formula transformation process

X*B0=LCM (x,b0) *GCD (x,b0);

X*B0=B1*GCD (X,B0);

X=B1/B0*GCD (X,B0);

Tidy up.

A1=GCD (X,A0);

X=B1/B0*GCD (X,B0);

First of all, it's very violent to think of enumeration x, so we can only get 50 points, so we need to optimize

Because GCD (x,b0) <=b0 and GCD (x,b0) are approximate b0, we enumerate gcd (x,b0) and gcd (x,b0) <=sqrt (B0);

Make I=GCD (x,b0) 1<=i<=sqrt (B0);

Determine if X=b1/b0*i and x=b1/b0* (b0/i) meet the conditions

And some special sentences, see the code.

#include <algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cstdio>#include<cmath>#include<queue>#definell Long Longusing namespacestd;Const intmaxn=1000000+10101; inlineintRead () {intx=0, f=1;CharCh=GetChar ();  for(;! IsDigit (CH); Ch=getchar ())if(ch=='-') f=-1;  for(; isdigit (ch); Ch=getchar ()) x= (x<<3) + (x<<1) +ch-'0'; returnx*F;}intn,a0,a1,b0,b1;intgcdintXinty) {    if(x<y) Swap (x, y); if(y==0)returnx; returnGCD (y,x%y); }intMain () {n=read ();  for(intI=1; i<=n;i++){        intans=0; A0=read (); A1=read (); B0=read (); b1=read (); intkk=sqrt (B0);  for(intI=1; i<=kk;i++){            if(b0%i!=0)Continue;//Note 1            intx1=b1/b0*i; if(GCD (x1,a0) ==a1 && gcd (x1,b0) ==i) ans++; if(b0/i==i)Continue;//NOTE 2x1=b1/b0* (b0/i); if(GCD (x1,a0) ==a1 && gcd (x1,b0) ==b0/i) ans++; } printf ("%d\n", ans); }    return 0;}

Hankson's Interesting questions