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
求p%m的b进制表示。果断java。
import java.io.*;
import java.math.*;
import java.util.*;
public class Main {
public static void main(String[] arges){
Scanner cin = new Scanner (System.in);
int b;
BigInteger p,m,ans;
while(cin.hasNext())
{
b=cin.nextInt();
String str;
if(b==0) break;
p=cin.nextBigInteger(b);
m=cin.nextBigInteger(b);
ans=p.mod(m);
str=ans.toString(b);
System.out.println(str);
}
}
}
POJ 2305 Basic remains(JAVA练习）
原文：http://blog.csdn.net/u013582254/article/details/38824245
