HDU 5050 Divided Land(最大公約數(shù)Java)
來源:程序員人生 發(fā)布時間:2014-10-12 18:52:16 閱讀次數(shù):2272次
題目鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=5050
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
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
PS:
思路很簡單,就是把輸入的二進制長和寬轉(zhuǎn)化為十進制求一個GCD然后在轉(zhuǎn)換為二進制輸出即可,不過由于數(shù)據(jù)過大,需要用Java來實現(xiàn),這里貼一發(fā)隊友敲的Java;
代碼如下:
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;
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);
a=gcd(a,b);
System.out.println("Case #"+i+": "+a.toString(2));
}
}
}
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