题目链接 http://codeforces.com/gym/101102/problem/C problem description Judge Bahosain was bored at ACM AmrahCPC 2016 as the winner of the contest had the first rank from the second hour until the end of the contest. Bahosain is studying the results of t…

Problem 1074: Hey Judge Time Limits:  1000 MS   Memory Limits:  65536 KB 64-bit interger IO format:  %lld   Java class name:  Main   Description Judge Nicole collected 7 ideas for problems of different levels, she wants to create 5 problems for the n…

转自:http://www.cnblogs.com/chouti/p/5752819.html Special Judge:当正确的输出结果不唯一的时候需要的自定义校验器 首先有个框架 #include<fstream> ifstream fin,fout,fstd ofstream fscore,freport; double Judge(){ } int main(int argc,char *argv[]) { //put something to fstreams... //Judge…

题目链接:http://acm.bnu.edu.cn/bnuoj/problem_show.php?pid=2418 解题报告:就是给你n个数,然后让你求这个数列的逆序对是多少?题目中n的范围是n < 500000,所以,暴力是不行的.还是第一次学会用线段树求逆序数,这种方法的时间复杂度是n * log n,是不是很快呢,利用了线段树查询速度快的优势.具体的方法如下: 这里先说一下,如果输入的n个数不是连续的,也就是说把这n个数按从小到大的顺序排列起来不是连续的话,还要先离散化一下,其实也就是把…

前段时间准备华为机试,正好之前看了一遍<剑指 Offer>,就在九度 Online Judge 上刷了书中的题目,使用的语言为 C++:只有3题没做,其他的都做了. 正如 Linus Torvalds 所言“Talk is cheap, show me the code!",详见托管在 Github 的源码:点击打开链接(核心部分有注释). 难度总结:有些可以不必要像书中那样保存中间结果,直接输出,也就可以不使用书中的算法:有些题目在书中题目的基础上加了更多的要求,也更难实现一点.…

留着当个模板用,在BNU上AC,在LA上RE……可能是java的提交方式不同??? 数和运算符各开一个栈. 表达式从左到右扫一遍,将数存成大数,遇到数压在 数的栈,运算符压在 运算符的栈,每当遇到右括号时,弹出 数的栈 的栈顶头两个元素,弹出 运算符的栈 顶的头一个元素,进行运算,将运算结果压回 数的栈 中.最后输出栈顶元素. 运算过程中把不符合情况的判掉. 我写的第二个java的题,竟然1A……这世界太不可思议了= = import java.util.*; import java.math.…

UVa Online Judge 工具網站   UVa中译题uHuntAlgorithmist Lucky貓的ACM園地,Lucky貓的 ACM 中譯題目 Mirror UVa Online Judge題目中譯!非常偉大的工作,請大家要心懷敬意! Ruby兔的ACM園地 UVa Online Judge題目中譯!非常偉大的工作,請大家要心懷感激! uHunt 這個站製作了一些網頁小工具, 可以查詢自己在UVa Online Judge的解題進度.簡單題列表.世界排名等等. 另外也整理了一套題庫,…

Submissions of online judge(1021) 问题描述 An online judge is a system to test programs in programming contests. The system can compile and execute codes, and test them with pre-constructed data. Submitted code may be run with restrictions, including tim…

Problem Description Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user's result file, then the system compare the two files. If the two f…

Big endian means the most significant byte stores first in memory. int a=0x01020304, if the cpu is big endian, data are store 01 02 03 04 in memory in increasing address. Below is a simple code to make the judgement. //execute the following cmd on th…