hdu 5050 java高精度GCD

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

310 100100 11010010 1100

Sample Output

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

题意：

二进制输入一个矩形的宽和长，然后让你分割成最大的正方形，问你最大正方形长度的二进制是多少。

思路：

如果不看二进制的话，给了长和宽找最大的分割正方形就是求一个长和宽的GCD。然而......2的1000次幂早已爆long  long,看来直接用GCD是妥妥的不行的。

那么接着考虑Java吧。

话说java真的是厉害啊，二进制可以直接转化为十进制，高精度也是库函数里的，默秒全C++.

import java.util.*;import java.math.BigInteger;import java.math.*;public class Main {public static void main(String[] args) {Scanner scanner=new Scanner(System.in);int t=scanner.nextInt();for(int cas=1;cas<=t;cas++){String str1=scanner.next();String str11=new BigInteger(str1,2).toString(10);BigInteger a=new BigInteger(str11);String str2=scanner.next();String str22=new BigInteger(str2,2).toString(10);BigInteger b=new BigInteger(str22);BigInteger ans=gcd(a,b);String str=ans.toString();String strr=new BigInteger(str,10).toString(2);System.out.print("Case #"+cas+": ");System.out.println(strr);}}static BigInteger gcd(BigInteger a,BigInteger b){BigInteger zero=new BigInteger("0");if(b.equals(zero))return a;return gcd(b,a.mod(b));}}