HDU Divided Land（Java大数，二进制大数最大公约数）

Divided Land

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1136    Accepted Submission(s): 485

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 ≤ 2¹⁰⁰⁰)

 

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

 

Source

2014 ACM/ICPC Asia Regional Shanghai Online

临近期末考试了，大部分时间用在复习上了，做题比较少了。
临近期末了，抽空学了学Java，发现Java用大数实在是太方便了。太适合这个题目了。

大体题意：
给你两个二进制大数，求出两个大数的最大公约数。
思路：
直接用Java   BigInteger  nextBigInteger（2）代表输入二进制大数。
并且自带gcd函数，输出在输出toString（2） 就是输出二进制大数了！

import java.math.BigInteger;import java.util.Scanner;public class Main {public static void main(String[] args){int T;Scanner sc = new Scanner(System.in);BigInteger a,b,c;T = sc.nextInt();for (int kase = 0; kase < T; ++kase){a = sc.nextBigInteger(2);b = sc.nextBigInteger(2);c = a.gcd(b);System.out.printf("Case #%d: ",kase+1);System.out.println(c.toString(2));}}}