HDOJ 1212 Big Number

Problem Description
As we know, Big Number is always troublesome. But it’s really important in our ACM. And today, your task is to write a program to calculate A mod B.

To make the problem easier, I promise that B will be smaller than 100000.

Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.

Input
The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.

Output
For each test case, you have to ouput the result of A mod B.

Sample Input
2 3
12 7
152455856554521 3250

Sample Output
2
5
1521

public BigDecimal remainder(BigDecimal divisor)返回其值为 (this % divisor) 的 BigDecimal。
余数由 this.subtract(this.divideToIntegralValue(divisor).multiply(divisor)) 给出。注意，这不是模操作（结果可以为负）。
参数：
divisor - 此 BigDecimal 要除以的值。
返回：
this % divisor。

import java.math.BigDecimal;
import java.util.Scanner;

public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); while(sc.hasNext()){ BigDecimal a = sc.nextBigDecimal(); int b = sc.nextInt(); System.out.println(a.remainder(new BigDecimal(b))); } }

}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/50649934