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。
代码语言：javascript复制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)));
}
}
}
