HDOJ 1212 Big Number

原创 陈 浩翔  2016-02-10 04:25  阅读 112 次

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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