大数求余（java高精度）

Link：http://poj.org/problem?id=2305

Problem：

Basic remains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4933		Accepted: 2081

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

import java.util.*;

import java.io.*;
import java.math.*;
import java.text.*;


public class Main{
public static void main(String args[])
{
BigInteger p,m,ans;
int b;
Scanner cin=new Scanner(System.in);
while(cin.hasNext())
{
b=cin.nextInt();
if(b==0)
break;
p=cin.nextBigInteger(b);
m=cin.nextBigInteger(b);
ans=p.mod(m);
String s;
s=ans.toString(b);
System.out.println(s);
}
}
}