一个重视算法和系统设计的实战派

【代码】高精度算法总结。

A + B Problem IIProblem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. InputThe first line of the input contain

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers

Notice that the number 123456789 is a 9-digit number consisting exactly the numbers from 1 to 9, with no duplication. Double it we will obtain 246913578, which happens to be another 9-digit number con

1005. Spell It Right (20)Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.Input Specification:Each i

题目描述： 输入一个正整数N，输出N的阶乘。输入：正整数N(0输出： 输入可能包括多组数据，对于每一组输入数据，输出N的阶乘样例输入：4515样例输出：241201307674368000来源：2006年清华大学计算机研究生机试真题答疑：解题遇到问题?分享解题心得?讨论本题请访问：htt

题目描述：实现一个加法器，使其能够输出a+b的值。输入：输入包括两个数a和b，其中a和b的位数不超过1000位。输出：可能有多组测试数据，对于每组数据，输出a+b的值。样例输入：2 610000000000000000000 10000000000000000000000000000000样例输出：810000

题目描述：    The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value

用Java写高精度题目真的很爽，发现Eclipse真的很强大，用Java编写代码的确很爽！只不过比较耗时，不过对提交题目够用了！import java.io.*;import java.util.*;import java.math.*;public class Main{ public static void main(String[] args){ Scanner in=ne

1. 输入：格式为：Scanner cin = new Scanner (new BufferedInputStream(System.in));例程：import java.io.*; import java.math.*; import java.util.*; import java.text.*; public class Main {

Chapter I. Java的优缺点各种书上都有，这里只说说用Java做ACM-ICPC的特点： (1) 最明显的好处是，学会Java，可以参加Java Challenge    :) (2) 对于熟悉C/C++的程序员来说，Java 并不难学，找本书，一两周业余时间就可以搞定了。当然，这里只是指一般编程，想熟悉所有的Java库还是需要些时间的。