1 题目 You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list. Input: (2 -> 4 -> 3) + (5 -> 6 -
在实际编码中,会遇到很多高精度的事例,比如,在计算金钱的时候就需要保留高精度小数,这样计算才不会有太大误差: 在下面的代码中,我们验证了,当两个float型的数字相加,得到的结果和我们的预期结果是有误差的,为了减小和防止这种误差的出现,我们需要使用BigInteger类和BigDecimal类来计算. package com.ietree.base.number; import java.math.BigDecimal; import java.math.BigInteger; public c
大数相加: package algorithm; //使用BigInteger类验证 import java.math.BigInteger; public class BigAdd { public static String bigNumberAdd(String f, String s) { // 翻转两个字符串,并转换成数组 char[] a = new StringBuffer(f).reverse().toString().toCharArray(); char[] b = new
Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2. Note: The length of both num1 and num2 is < 5100. Both num1 and num2 contains only digits 0-9. Both num1 and num2 does not contain any leading zero.
题目 我要开始练习一些java的简单编程了^v^ import java.io.*; import java.util.*; import java.math.*; public class Main { /** * @param args */ public static void main(String[] args) throws Exception { // 定义并打开输入文件 Scanner cin = new Scanner(System.in); BigInteger a, sum
我们可以把一个很大很长的数分成多个短小的数,然后保存在一个数组中,大数之间的四则运算及其它运算都是通过数组完成.JDK就是这么实现的.JDK的BigInteger类里用一个int数组来保存数据: /** * The magnitude of this BigInteger, in <i>big-endian</i> order: the * zeroth element of this array is the most-significant int of the * magni
A + B Problem II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 313689 Accepted Submission(s): 60742 Problem Description I have a very simple problem for you. Given two integers A and B, your job
Segment Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 5666 Problem Description Silen August does not like to talk with others.She like to find some interesting problems. Today she find