POJ 2305 Basic remains（大数取模）（JAVA）

Basic remains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4076		Accepted: 1707

Description

Given a base b and two non-negative base b integers p and m, compute p mod m and print the result as a base b integer. p mod m is defined as the smallest non-negative integer k such that p = a*m + k for some integer a.

Input

Input consists of a number of cases. Each case is represented by a line containing three unsigned integers. The first, b, is a decimal number between 2 and 10. The second, p, contains up to 1000 digits between 0 and b-1. The third, m, contains up to 9 digits between 0 and b-1. The last case is followed by a line containing 0.

Output

For each test case, print a line giving p mod m as a base-b integer.

Sample Input

2 1100 101

10 123456789123456789123456789 1000

0

Sample Output

10

789

Source

Waterloo local 2003.09.20

 解题报告：以b进制输入p和m，求p模m的b进制结果 ，用Java写挺简单的

代码如下：

import java.util.Scanner;
import java.math.BigInteger;
public class Main{
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        BigInteger ans, p, m;
        int b;
        String str;
        while (scan.hasNextInt()){
            b = scan.nextInt();
            if (b == 0){
                break;
            }
            p = scan.nextBigInteger(b);
            m = scan.nextBigInteger(b);
            ans = p.mod(m);
            str = ans.toString(b);//将ans转化成b进制形式
            System.out.println(str);
        }
    }
}