link:http://acm.hdu.edu.cn/showproblem.php?pid=4690

考的是耐心何细心啊,用map把两个表格映射一下就行~

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <cmath>
 #include <cctype>
 #include <algorithm>
 #include <queue>
 #include <deque>
 #include <queue>
 #include <list>
 #include <map>
 #include <set>
 #include <vector>
 #include <utility>
 #include <functional>
 #include <fstream>
 #include <iomanip>
 #include <sstream>
 #include <numeric>
 #include <cassert>
 #include <ctime>
 #include <iterator>
 const int INF = 0x3f3f3f3f;
 ][] = {{-,},{,},{,-},{,},{-,-},{-,},{,-},{,}};
 using namespace std;
 ][] = {
     "NUL","SOH","STX","ETX","","HT","","DEL","","","","VT","FF","CR","SO","SI",
     "DLE","DC1","DC2","DC3","","","BS","","CAN","EM","","","IFS","IGS","IRS","IUS ITB",
     "","","","","","LF","ETB","ESC","","","","","","ENQ","ACK","BEL",
     "","","SYN","","","","","EOT","","","","","DC4","NAK","","SUB",
     "SP","","","","","","","","","","",".","<","(","+","|",
     "&","","","","","","","","","","!","$","*",")",";","",
     "-","/","","","","","","","","","",",","%","_",">","?",
     "","","","","","","","","","`",":","#","@","'","=","\"",
     "","a","b","c","d","e","f","g","h","i","","","","","","",
     "","j","k","l","m","n","o","p","q","r","","","","","","",
     "","~","s","t","u","v","w","x","y","z","","","","","","",
     "^","","","","","","","","","","[","]","","","","",
     "{","A","B","C","D","E","F","G","H","I","","","","","","",
     "}","J","K","L","M","N","O","P","Q","R","","","","","","",
     "\\","","S","T","U","V","W","X","Y","Z","","","","","","",
     ","","","","","",""
 };
 ][] = {
 "NUL",    "SOH",    "STX",    "ETX",    "EOT",    "ENQ",    "ACK",    "BEL",    "BS",    "HT",    "LF",    "VT",    "FF",    "CR",    "SO",    "SI",
 "DLE",    "DC1",    "DC2",    "DC3",    "DC4",    "NAK",    "SYN",    "ETB",    "CAN",    "EM",    "SUB",    "ESC",    "IFS",    "IGS",    "IRS",    "IUS ITB",
 "SP",    "!",    "\"",    "#",    "$",    "%",    "&",    "'",    "(",    ")",    "*",    "+",    ",",    "-",    ".",    "/",
 ",    ":",    ";",    "<",    "=",    ">",    "?",
 "@",    "A",    "B",    "C",    "D",    "E",    "F",    "G",    "H",    "I",    "J",    "K",    "L",    "M",    "N",    "O",
 "P",    "Q",    "R",    "S",    "T",    "U",    "V",    "W",    "X",    "Y",    "Z",    "[", "\\","]","^","_",
 "`",    "a",    "b",    "c",    "d",    "e",    "f",    "g",    "h",    "i",    "j",    "k",    "l",    "m",    "n",    "o",
 "p",    "q",    "r",    "s",    "t",    "u",    "v",    "w",    "x",    "y",    "z",    "{",    "|",    "}",    "~",    "DEL"
 };
 map<pair<int,int>, pair<int,int> > coll;
 int main(void)
 {
     #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin );
     #endif // ONLINE_JUDGE
     string sad;
     ios::sync_with_stdio(false);
     cin>>sad; coll.clear();
     string tmp; pair<int,int> so, to;
     ; i < ; ++i)
     {
         ; j < ; ++j)
         {
             if (A[i][j]!="")
             {
                 so.first=i, so.second=j;
                 bool mrk = false;
                 ; k < ; ++k)
                 {
                     ; h < ; ++h)
                     {
                         if (A[i][j]==B[k][h])
                         {
                             mrk = true;
                             to.first=k,to.second=h;
                             coll[so] = to; break;
                         }
                     }
                     if (mrk) break;
                 }
             }
         }
     }
     ; i < sad.size(); i+=)
     {
         int X, Y; tmp.clear();
         ;
         ';
         ]>=]<=]- ;
         ] - ';
         pair<int,int> hehe, TM;
         hehe.first = X, hehe.second = Y;
         TM = coll[hehe];
         int j = TM.first, k = TM.second;
         char XX, YY;
         XX = j+ ';
         ) YY = k -  + 'A';
         ';
         printf("%c%c", XX, YY);
     }
     printf("\n");

     ;
 }

写那两个数组没有vim的帮助手是不是得残了……

还有一个就是,当你发现程序某一小段怎么都不对的时候,也许重新写一下就OK了,调试了很久……

走吧,小胖!

hdu4690 EBCDIC ——水题,考耐心的更多相关文章

  1. 【转】POJ百道水题列表

    以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight ...

  2. HDOJ 2317. Nasty Hacks 模拟水题

    Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Tota ...

  3. ACM :漫漫上学路 -DP -水题

    CSU 1772 漫漫上学路 Time Limit: 1000MS   Memory Limit: 131072KB   64bit IO Format: %lld & %llu Submit ...

  4. ytu 1050:写一个函数,使给定的一个二维数组(3×3)转置,即行列互换(水题)

    1050: 写一个函数,使给定的一个二维数组(3×3)转置,即行列互换 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 154  Solved: 112[ ...

  5. [poj2247] Humble Numbers (DP水题)

    DP 水题 Description A number whose only prime factors are 2,3,5 or 7 is called a humble number. The se ...

  6. gdutcode 1195: 相信我这是水题 GDUT中有个风云人物pigofzhou,是冰点奇迹队的主代码手,

    1195: 相信我这是水题 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 821  Solved: 219 Description GDUT中有个风云人 ...

  7. BZOJ 1303 CQOI2009 中位数图 水题

    1303: [CQOI2009]中位数图 Time Limit: 1 Sec  Memory Limit: 162 MBSubmit: 2340  Solved: 1464[Submit][Statu ...

  8. 第十一届“蓝狐网络杯”湖南省大学生计算机程序设计竞赛 B - 大还是小? 字符串水题

    B - 大还是小? Time Limit:5000MS     Memory Limit:65535KB     64bit IO Format: Description 输入两个实数,判断第一个数大 ...

  9. ACM水题

    ACM小白...非常费劲儿的学习中,我觉得目前我能做出来的都可以划分在水题的范围中...不断做,不断总结,随时更新 POJ: 1004 Financial Management 求平均值 杭电OJ: ...

随机推荐

  1. 访问google.com

    ping www.google.com 得到的IP来访问

  2. 原生Ajax讲解

    典型的http通信:浏览器向服务器发出请求,服务器向客户端返回响应,浏览器重新加载页面,这种不连续的页面加载方式导致用户的体验变得杂乱,缺乏连贯性. 如: 在一般的web应用程序中,用户填写表单字段然 ...

  3. 邮箱输入(仿gmail)

    年前同事做邮件,我调研了几个如163.qq等的邮箱,最终觉得还是gmail的用着舒服,看着也舒服.就仿照写了个.还有问题.记录下,有时间再整理下代码. demo

  4. Java线程中yield与join方法的区别

    长期以来,多线程问题颇为受到面试官的青睐.虽然我个人认为我们当中很少有人能真正获得机会开发复杂的多线程应用(在过去的七年中,我得到了一个机会),但是理解多线程对增加你的信心很有用.之前,我讨论了一个w ...

  5. 如何让linux定时任务crontab按秒执行

    如何让linux定时任务crontab按秒执行? linux定时任务crontab最小执行时间单位为分钟如果想以秒为单位执行,应该如何设置呢?思路 正常情况是在crontab中直接定义要执行的任务,现 ...

  6. Ubuntu中设置静态IP和DNS

    在Ubuntu中设置静态IP共两步:1>设置IP:2>设置DNS1>设置IP    编辑 /etc/network/interface文件:       sudo vi /etc/n ...

  7. winform app.config文件的动态配置

    获取 获取应用程序exe.config文件中  节点value值 /// <summary> /// 功能: 读取应用程序exe.config文件中 /// appSettings节点下 ...

  8. Educational Codeforces Round 14 D. Swaps in Permutation

    题目链接 分析:一些边把各个节点连接成了一颗颗树.因为每棵树上的边可以走任意次,所以不难想出要字典序最大,就是每棵树中数字大的放在树中节点编号比较小的位置. 我用了极为暴力的方法,先dfs每棵树,再用 ...

  9. JS判断form内所有表单是否为空

    function checkForm(){ var input_cart=document.getElementsByTagName("INPUT"); for(var   i=0 ...

  10. 学习swift开源项目

    如果你是位iOS开发者,或者你正想进入该行业,那么Swift为你提供了一个绝佳的机会.Swift的设计非常优雅,较Obj-C更易于学习,当然也非常强大. 为了指导开发者使用Swift进行开发,苹果发布 ...