java大整数求余_大数求余（java高精度）

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#java大整数求余

本文介绍了一种解决大数取模问题的方法，通过给定的基数b和两个非负基数b整数p和m，计算p mod m的结果，并以基数b的形式输出。文章提供了一个Java实现案例，演示了如何读取输入并使用BigInteger类来处理大数运算。

Problem：

Basic remains

Time Limit: 1000MS

Memory Limit: 65536K

Total Submissions: 4933

Accepted: 2081

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

Sample Output

789

Source

import java.util.*; import java.io.*; import java.math.*; import java.text.*; public class Main{ public static void main(String args[]) { BigInteger p,m,ans; int b; Scanner cin=new Scanner(System.in); while(cin.hasNext()) { b=cin.nextInt(); if(b==0) break; p=cin.nextBigInteger(b); m=cin.nextBigInteger(b); ans=p.mod(m); String s; s=ans.toString(b); System.out.println(s); } } }