HDU 5050 Divided Land （二进制上的最大公约数）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5050

题面：

Divided Land

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1024    Accepted Submission(s): 438

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

   
   
   
   

    
    
    
    3
10 100
100 110
10010 1100
   
   
   
   

 

Sample Output

   
   
   
   

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

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online

 

题目大意：

    求两个以二进制形式给出的数的最大公约数的二进制表示。

解题：

   这是有一条公式的，不知道可能就没法做。

   1.a，b都为偶数。gcd(a,b)=2*gcd(a/2,b/2)。

   2.a为偶数，b为奇数。gcd(a,b)=gcd(a/2,b)。

   3.a,b都为奇数（设a大于等于b）。gcd(a,b)=gcd((a-b),b)。

然后用字符串模拟即可。（代码比较挫，还是自己写吧，囧）

代码：

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
string ans;
int cnt;
//比较大小
bool compare(string x,string y)
{  return x>y;}
//取小
string min(string x,string y)
{
	if(x.length()<y.length())
		return x;
    else if(x.length()>y.length())
		return y;
	else 
	{
		if(compare(x,y))
			return y;
		else return x;
	}
}
//取大
string max(string x,string y)
{
	if(x.length()>y.length())
		return x;
	else if(x.length()<y.length())
		return y;
	else
	{
		if(compare(x,y))
			return x;
		else return y;
	}
}
//相减
string subtract(string x,string y)
{
	string res="";
	reverse(x.begin(),x.end());
	reverse(y.begin(),y.end());
	y+=string(x.length()-y.length(),'0');
	int len=x.length(),bor=0,tmp;
	for(int i=0;i<len;i++)
	{
		tmp=x[i]-y[i]+bor;
		if(tmp==0)
		{
			res+='0';
			bor=0;
		}
		else if(tmp==1)
		{
			res+='1';
			bor=0;
		}
		else if(tmp==-1)
		{
			res+='1';
			bor=-1;
		}
		else
		{
			res+='0';
			bor=-1;
		}
	}
	reverse(res.begin(),res.end());
	bool sign=true;
	len=res.length();
	int p=-1;
	for(int i=0;i<len;i++)
	{
			if(res[i]=='0')
				p=i;
			else
				break;
	}
	//去前导0
	res=res.substr(p+1);
	return res;
}
//求公约数
void gcd(string x,string y)
{
	string minn,maxn; 
	//y已经为0了
	if(y=="")
	{
		ans=x;
		return;
	}
	char s,t;
	s=x[x.length()-1];
	t=y[y.length()-1];
	//同偶
	if(s=='0'&&t=='0')
	{
		cnt++;
		gcd(x.substr(0,x.length()-1),y.substr(0,y.length()-1));
	}
	//同奇
	else if(s=='1'&&t=='1')
	{
        minn=min(x,y);
        maxn=max(x,y);
		gcd(minn,subtract(maxn,minn));
	}
	//一奇一偶
	else if(s=='0'&&t=='1')
	{
	   gcd(x.substr(0,x.length()-1),y);
	}
	else 
	   gcd(x,y.substr(0,y.length()-1));
}
int main()
{
	int t;
	cin>>t;
    string a,b;
	for(int i=1;i<=t;i++)
	{
	   cnt=0;
	   cin>>a>>b;
	   gcd(a,b);
	   //补约掉的2
	   for(int j=0;j<cnt;j++)
	   ans+='0';
	   cout<<"Case #"<<i<<": ";
	   cout<<ans<<endl;
	}
	return 0;
}