Algorithm
problem Link : https://onlinejudge.org/index.php?option=onlinejudge&Itemid=8&page=show_problem&problem=1492
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
Code Examples
#1 Code Example with C Programming
Code -
C Programming
import java.util.*;
import java.math.BigInteger;
public class Main {
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
while(true){
int base = sc.nextInt();
if(base == 0) break;
BigInteger p = new BigInteger(sc.next(),base); // convert base x string to base 10
BigInteger m = new BigInteger(sc.next(),base);
p = p.mod(m);
System.out.println(p.toString(base)); // output as base x
}
}
}
Input
10 123456789123456789123456789 1000
0
Output
789
Demonstration
UVA Online Judge solution - 10551-Basic Remains.java - UVA Online Judge solution in C,C++,java