HDU-5050 divided Land (binary seeking GCD)

The main topic: The GCD of two binary numbers is represented by a binary number.

Topic Analysis: This problem can be used in Java in the large number of AC.

The code is as follows:

Import Java.io*;import Java.math.biginteger;import Java.util.scanner;public class Main {public static void main (String Agrs[]) {Scanner sc=new Scanner (system.in); int t=sc.nextint (); for (int i=1;i<=t;++i) {String p=sc.next (); String Q=sc.next (); BigInteger a=new BigInteger (p,2); BigInteger b=new BigInteger (q,2); A=A.GCD (b); System.out.print ("Case #" +i+ ":"); System.out.println (a.tostring (2));} Sc.close ();}}

　　

However, you can also use binary to find GCD.

GCD (A, B) =gcd (A/2,B/2) * (A, b all are even)

GCD (b) =gcd (A,B/2) (A is odd, B is even)

GCD (A, B) =gcd ((a)/2,b) (A, b are odd)

Unfortunately I use C + + without AC, here is my code without AC:

# include<iostream># include<cstdio># include<cstring># include<algorithm># include<    String>using namespace Std;bool isbigger (string p,string q) {if (P.length () >q.length ()) return true;    else if (P.length () <q.length ()) return false;        else{int len=p.length ();            for (int i=len-1;i>=0;--i) {if (P[i]>q[i]) return true;        else if (P[i]<q[i]) return false;    }}}string Sub (string p,string q) {int len=q.length ();        for (int i=0;i<len;++i) {if (P[i]>=q[i]) p[i]=p[i]-q[i]+ ' 0 ';            else{p[i+1]-=1;        p[i]=p[i]+2-q[i]+ ' 0 ';            }} for (int i=len;i<p.length (); ++i) {if (p[i]< ' 0 ' &&i+1<p.length ()) {--p[i+1];        p[i]+=2;    }} len=p.length ();    while (p[len-1]<= ' 0 ' &&len>0)--len; Return P.substr (0,len);}    String f (String p,string Q) {if (p==q) return p;    if (p== "1") return "1"; If(q== "1") return "1";    int N=p.length (), m=q.length ();    if (p[0]== ' 0 ' &&q[0]== ' 0 ') return F (P.substr (1,n-1), Q.substr (1,m-1)) + ' 0 ';    else if (p[0]== ' 1 ' &&q[0]== ' 0 ') return F (P,q.substr (1,m-1));    else if (p[0]== ' 0 ' &&q[0]== ' 1 ') return F (P.substr (1,n-1), q);            else{if (Isbigger (p,q)) {p=sub (p,q);            N=p.length ();        Return f (P.substr (1,n-1), q);            }else{q=sub (q,p);            M=q.length ();        Return f (P,q.substr (1,m-1));    }}}int Main () {int T;    String p,q;    scanf ("%d", &t);        for (int i=1;i<=t;++i) {cin>>p>>q;        Reverse (P.begin (), P.end ());        Reverse (Q.begin (), Q.end ());    cout<< "Case #" <<i<< ":" <<f (P,Q) <<endl; } return 0;}

　　

HDU-5050 divided Land (binary seeking GCD)