POJ 2305

Basic remains

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4429 Accepted: 1869

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 10110 123456789123456789123456789 10000

Sample Output

10789

Source

Waterloo local 2003.09.20

训练下java的大数，来自 MMchen

题意：在 b 进制下，算出 p mod m

2305Accepted5916K829MSJava527B2013-05-09 11:35:20 

import java.math.*;import java.util.*;public class Main {public static void main(String args[]){int base;Scanner cin = new Scanner(System.in);while(cin.hasNextInt()){base = cin.nextInt();if(base == 0) break;BigInteger p, m;p = cin.nextBigInteger(base); //自动转换为 base进制读入m = cin.nextBigInteger(base);BigInteger a;String str;a = p.mod(m); // a = p%m 注意此时结果为 10 进制str = a.toString(base); //将十进制 a 转换成  base 进制字符串System.out.println(str);}}}

         

2305Accepted5916K829MSJava527B2013-05-09 11:35:20