JAVA大数求余 Basic remains(POJ 2305)

最新推荐文章于 2025-12-08 09:18:56 发布

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本文介绍了一种使用Java进行大数运算的方法，通过实例演示了如何计算两个大数p和m之间的模运算p mod m，并展示了解决方案的代码实现。

题目描述

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.

输入

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.

输出

For each test case, print a line giving p mod m as a base-b integer.

样例输入

2 1100 101
10 123456789123456789123456789 1000
0

样例输出

import java.math.BigInteger;
import java.util.Scanner;
public class Main3 {
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		while (scanner.hasNext()) {
			int n = scanner.nextInt();
			if (n == 0)
				break;
			BigInteger aBigInteger = scanner.nextBigInteger();
			BigInteger bBigInteger = scanner.nextBigInteger();
			BigInteger a = new BigInteger(aBigInteger.toString(), n);
			BigInteger b = new BigInteger(bBigInteger.toString(), n);
			BigInteger cBigInteger = a.divide(b);
			BigInteger mBigInteger = a.subtract(cBigInteger.multiply(b));
			System.out.println(mBigInteger.toString(n));
		}
	}
}