hdu 5050 Divided Land

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#二进制 #java #最大公约数

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本文探讨了一个关于二进制数的最大公约数问题，通过两种不同的编程语言实现了解决方案：一种是使用C++实现的更相减数法，另一种是使用Java实现的基于BigInteger的算法。最终目标是在给定的二进制长度和宽度的情况下，找到能够平均分割矩形土地的最大二进制宽度。

Divided Land

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 402 Accepted Submission(s): 213

Problem Description

It’s time to fight the local despots and redistribute the land. There is a rectangular piece of land granted from the government, whose length and width are both in binary form. As the mayor, you must segment the land into multiple squares of equal size for the villagers. What are required is there must be no any waste and each single segmented square land has as large area as possible. The width of the segmented square land is also binary.

Input

The first line of the input is T (1 ≤ T ≤ 100), which stands for the number of test cases you need to solve.

Each case contains two binary number represents the length L and the width W of given land. (0 < L, W ≤ 2 ¹⁰⁰⁰)

Output

For each test case, print a line “Case #t: ”(without quotes, t means the index of the test case) at the beginning. Then one number means the largest width of land that can be divided from input data. And it will be show in binary. Do not have any useless number or space.

Sample Input

Sample Output

  
   Case #1: 10
Case #2: 10
Case #3: 110

Source

2014 ACM/ICPC Asia Regional Shanghai Online

这个题意是两个二进制数求他的最大公约数。

我用C++想到的方法就只有更相相减数，结果超时了。所以只能用java试试。据说C++有很好的方法，下次学习下。mark！

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
#include <ctime>
#define LL __int64
#define EPS 1e-8
using namespace std;
string s1,s2;
int pd(string a,string b)
{
	int l1=a.length();
	int l2=b.length();
	if (l1<l2) return 1;
	if (l1>l2) return 0;
	for (int i=0;i<l1;i++)
	{
		if (s1[i]>s2[i]) return 0;
		if (s1[i]<s2[i]) return 1;
	}	
}
void go(string a,string b)
{
	int l=b.length();
	int l1=a.length()-1;
	for (int i=l-1;i>=0;i--)
	{
		a[l1]='0'+a[l1]-b[i];
	//	cout<<a<<"@"<<i<<endl;
		if (a[l1]=='/')
		{
			//cout<<"@@@"<<endl;
			int j=l1;
			while (a[j]=='/')
			{
				a[j]='1';
				a[j-1]='0'+a[j-1]-b[i];
				j--;
			}
		}
		l1--;
	}
	int i=0;
	while (a[i]=='0') i++;
	//cout<<a<<endl;
	a.assign(a,i,a.length()-i);
	//cout<<a<<"*"<<b<<endl;
	int g=pd(a,b);
	if (g==0)
	{
		s1=a;
		s2=b;
	}
	else
	{
		s1=b;
		s2=a;
	}
}
int main()
{
	int l1,l2,T,cas=1;
	scanf("%d",&T);
	while (T--)
	{
		getchar();
		cin>>s1>>s2;
		l1=s1.length();
		l2=s2.length();
		//cout<<s1<<" "<<s2<<endl;
		while (s1!=s2 && s2!="" && s1!="")
		{
			int f=pd(s1,s2);
			if (f) go(s2,s1);
			else go(s1,s2);
			//cout<<s1<<" "<<s2<<endl;
		}
		printf("Case #%d: ",cas++);
		cout<<s1<<endl;
	}
	return 0;
}

package hdu5050;

import java.math.*;
import java.util.Scanner;

public class Main{
	public static BigInteger gcd(BigInteger a,BigInteger b)
	{
		if (b.equals(BigInteger.ZERO))
			return a;
		else return gcd(b,a.mod(b));
	}
	public static void main(String[] args)
	{
		Scanner input=new Scanner(System.in);
		int t,i,j;
		String s=null;
		char str[];
		BigInteger a,b;
		t=input.nextInt();
		for (i=1;i<=t;i++)
		{
			a=input.nextBigInteger(2);
			b=input.nextBigInteger(2);
			System.out.println(a);
			a=gcd(a,b);
			System.out.println("Case #"+i+": "+a.toString(2));
		}
	}
}