摘要: The Last Non-zero DigitTime Limit:1000MSMemory Limit:65536KTotal Submissions:4259Accepted:1205DescriptionIn this problem you will be given two decimal integer number N, M. You will have to find the last non-zero digit of theNPM.This means no of permutations of N things taking M at a time.InputThe in 阅读全文
posted @ 2013-03-30 18:25 Jack Ge 阅读(664) 评论(0) 推荐(0) 编辑
摘要: Legal or NotTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2672Accepted Submission(s): 1191 Problem DescriptionACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many & 阅读全文
posted @ 2013-03-30 15:31 Jack Ge 阅读(487) 评论(0) 推荐(0) 编辑
摘要: (转)http://www.cnblogs.com/XBWer/archive/2012/06/24/2560532.htmlACM中java的使用这里指的java速成,只限于java语法,包括输入输出,运算处理,字符串和高精度的处理,进制之间的转换等,能解决OJ上的一些高精度题目。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 c 阅读全文
posted @ 2013-03-30 10:34 Jack Ge 阅读(6997) 评论(0) 推荐(2) 编辑
摘要: A + B Problem IITime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 149365Accepted Submission(s): 28157 Problem 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 阅读全文
posted @ 2013-03-30 10:21 Jack Ge 阅读(309) 评论(0) 推荐(0) 编辑
摘要: Hat's FibonacciTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4547Accepted Submission(s): 1544 Problem DescriptionA Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1. F(1 阅读全文
posted @ 2013-03-30 10:15 Jack Ge 阅读(330) 评论(0) 推荐(0) 编辑
摘要: N!Time Limit: 10000/5000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 40204Accepted Submission(s): 11162 Problem DescriptionGiven an integer N(0 ≤ N ≤ 10000), your task is to calculate N!InputOne N in one line, process to the end of file.OutputFor each N, output N! 阅读全文
posted @ 2013-03-30 10:08 Jack Ge 阅读(170) 评论(0) 推荐(0) 编辑
摘要: Children’s QueueTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8194Accepted Submission(s): 2603 Problem DescriptionThere are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He pres 阅读全文
posted @ 2013-03-30 10:01 Jack Ge 阅读(573) 评论(0) 推荐(0) 编辑
摘要: 大菲波数Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7677Accepted Submission(s): 2557 Problem DescriptionFibonacci数列,定义如下: f(1)=f(2)=1 f(n)=f(n-1)+f(n-2) n>=3。 计算第n项Fibonacci数值。Input输入第一行为一个整数N,接下来N行为整数Pi(1<=Pi<=1000)。Output输出为N行,每行为对应的f(P 阅读全文
posted @ 2013-03-30 09:44 Jack Ge 阅读(213) 评论(0) 推荐(0) 编辑
摘要: 大明A+BTime Limit: 3000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5747Accepted Submission(s): 1957 Problem Description话说,经过了漫长的一个多月,小明已经成长了许多,所以他改了一个名字叫“大明”。 这时他已经不是那个只会做100以内加法的那个“小明”了,现在他甚至会任意长度的正小数的加法。现在,给你两个正的小数A和B,你的任务是代表大明计算出A+B的值。Input本题目包含多组测试数据,请处理到文件 阅读全文
posted @ 2013-03-30 09:35 Jack Ge 阅读(200) 评论(0) 推荐(0) 编辑
摘要: 字串数Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2142Accepted Submission(s): 458 Problem Description一个A和两个B一共可以组成三种字符串:"ABB","BAB","BBA". 给定若干字母和它们相应的个数,计算一共可以组成多少个不同的字符串.Input每组测试数据分两行,第一行为n(1<=n<=26),表示不同字母的 阅读全文
posted @ 2013-03-30 09:23 Jack Ge 阅读(1787) 评论(0) 推荐(0) 编辑
摘要: ExponentiationTime Limit: 1000/500 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4973Accepted Submission(s): 1353 Problem DescriptionProblems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of t 阅读全文
posted @ 2013-03-30 08:45 Jack Ge 阅读(492) 评论(0) 推荐(0) 编辑
摘要: %ALLUSERSPROFILE% : 列出所有用户Profile文件位置。%APPDATA% : 列出应用程序数据的默认存放位置。%CD% : 列出当前目录。%CLIENTNAME% : 列出联接到终端服务会话时客户端的NETBIOS名。%CMDCMDLINE% : 列出启动当前cmd.exe所使用的命令行。%CMDEXTVERSION% : 命令出当前命令处理程序扩展版本号。%CommonProgramFiles% : 列出了常用文件的文件夹路径。%COMPUTERNAME% : 列出了计算机名。%COMSPEC% : 列出了可执行命令外壳(命令处理程序)的路径。%DATE% : 列出当前 阅读全文
posted @ 2013-03-30 08:33 Jack Ge 阅读(320) 评论(0) 推荐(0) 编辑