JAVA大数求余 Basic remains(POJ 2305)

Basic remains(POJ 2305)

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2863		Accepted: 1169

DescriptionGiven 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.InputInput 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.OutputFor 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

CODE:


//
import java.util.*;
import java.math.*;
import java.io.*;
public class HDOJ_1000 {
public static void main(String[] args)
{
BigInteger p,m,mul,N;
int i,len;
int n;
String  a,b;

Scanner cin=new Scanner(System.in);
while(cin.hasNext())
{

n=cin.nextInt();
if(n==0)
break;
p=BigInteger.valueOf(0);
m=BigInteger.valueOf(0);
a=cin.next();
b=cin.next();
len=a.length();
mul=BigInteger.valueOf(1);

N=BigInteger.valueOf(n);

for(i=len-1;i>=0;i--)
{
p=p.add(BigInteger.valueOf((a.charAt(i)-'0')).multiply(mul));
mul=mul.multiply(N);
}
len=b.length();
mul=BigInteger.valueOf(1);
for(i=len-1;i>=0;i--)
{
m=m.add(BigInteger.valueOf((b.charAt(i)-'0')).multiply(mul));
mul=mul.multiply(N);

}
System.out.println(p.mod(m).toString(n));

}
}

}

//