题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1073 数据判断,主要是如何判断WA和PE,吸收字符! 代码: #include <stdio.h> #include <string.h> #include <math.h> #include <algorithm> #include <iostream> #include <ctype.h> #include <iomanip&g…

Online Judge Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6534    Accepted Submission(s): 2472 Problem Description Ignatius is building an Online Judge, now he has worked out all the problem…

对于开始学习C语言程序设计或C++程序设计面向过程部分的同学来说,利用在线OJ网站进行实践训练,对提高自己的编程能力很有好处.国内外OJ网站很多,每个都去看看,去刷个题,是不现实的,也没必要.即使一个OJ网站,上面3~4千道题也难全部刷完.因此,给大家推荐两个OJ网站.一个是北京大学的PKU JudgeOnline(http://poj.org/),简称POJ:另一个是杭州电子科技大学的HDU Online Judge System (http://acm.hdu.edu.cn/),简称 HDU…

Online Judge系统      Online Judge系统(简称OJ)是一个在线的判题系统.用户可以在线提交给定问题的多种程序(如C.C++.Pascal.Java)源代码,系统对源代码进行编译和执行,并通过预先设计的测试数据来检验程序源代码的正确性. 一个用户提交的程序在Online Judge系统下执行时将受到比较严格的限制,包括运行时间限制.内存使用限制和安全限制等.用户程序执行的结果将被Online Judge系统捕捉并保存,然后再转交给一个裁判程序.该裁判程序或者比较用户程序…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1073 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,…

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…

题目链接 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…

Online Judge Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4977    Accepted Submission(s): 1889 Problem Description Ignatius is building an Online Judge, now he has worked out all the problem…

模拟评测机判断答案 先判断有没有不一样的 有的话再提取出 有效子列 看看有没有错的 #include <iostream> #include <cstdio> #include <cstring> #include <string> using namespace std; int t; string a,b,sa,sb; ]; ][]={"Accepted","Presentation Error","Wr…

//数据是有多水 连 10^10的枚举都能过 关于拓展欧几里德:大概就是x1=y2,y1=x2-[a/b]y2,按这个规律递归到gcd(a,0)的形式,此时公因数为a,方程也变为a*x+0*y=gcd(a,0)的形式,显然解为x=1,y=0,然后再递归回去就能得到解(a*x+b*y=gcd(a,b)的解) #include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #incl…