HUST 1358 Uiwurerirexb jeqvad(模拟解密)

Description

Fmur lan oxbrvu mzx, E abpxcay Jvmffabza qdxwfaou eb vmjsad.xdz, eb nvejv rvada mda ombc jdcqrxzdmqvc qdxwfaou. Moxbz rvaua qdxwfaou rvada mda ombc iuebz uiwurerirexb jeqvad oarvxy. Rx sbxn oxda mwxir er, E nabr rx Neseqayem:
Eb jdcqrxzdmqvc, m uiwurerirexb jeqvad eu m oarvxy xl abjdcqrexb wc nvejv iberu xl qfmebrakr mda daqfmjay nerv jeqvadrakr mjjxdyebz rx m dazifmd ucurao; rva "iberu" omc wa uebzfa farradu (rva oxur jxooxb), qmedu xl farradu, rdeqfaru xl farradu, oekridau xl rva mwxta, mby ux lxdrv. Rva dajaetad yajeqvadu rva rakr wc qadlxdoebz mb ebtadua uiwurerirexb. Uiwurerirexb jeqvadu jmb wa jxoqmday nerv rdmbuqxuerexb jeqvadu. Eb m rdmbuqxuerexb jeqvad, rva iberu xl rva qfmebrakr mda damddmbzay eb m yelladabr mby iuimffc giera jxoqfak xdyad, wir rva iberu rvaouaftau mda falr ibjvmbzay. Wc jxbrdmur, eb m uiwurerirexb jeqvad, rva iberu xl rva qfmebrakr mda darmebay eb rva umoa uagiabja eb rva jeqvadrakr, wir rva iberu rvaouaftau mda mfraday. Rvada mda m biowad xl yelladabr rcqau xl uiwurerirexb jeqvad. El rva jeqvad xqadmrau xb uebzfa farradu, er eu radoay m ueoqfa uiwurerirexb jeqvad; m jeqvad rvmr xqadmrau xb fmdzad zdxiqu xl farradu eu radoay qxfczdmqvej. Moxbxmfqvmwarej jeqvad iuau lekay uiwurerirexb xtad rva abreda oauumza, nvadamu m qxfcmfqvmwarej jeqvadiuau m biowad xl uiwurerirexbu mr yelladabr reoau eb rva oauumza, nvada m iber ldxo rva qfmebrakr eu omqqay rx xba xl uatadmf qxuuewefereau eb rva Jeqvadrakr mby teja-tadum.
Rveu reoa E zeta cxi uijv m qdxwfao, eb nvejv Yaujdeqrexb, Ebqir, Xirqir, Umoqfa Ebqir mda mff jeqvadrakr, nvadamu Umoqfa Xirqir eu iuebz qfmebrakr. Xl jxidua, rveu qdxwfao eu iuebz ueoqfa uiwurerirexb jeqvad mby oxbxmfqvmwarej jeqvad. Rx uxfta oxbxmfqvmwarej jeqvad allejeabrfc, cxi jmb dalad rxvrrq://nnn.uajdarjxyawdamsad.jxo/UJWUxftd.heq

Input

Rva ebqir jxbrmebu oifreqfa jmuau. Amjv jmua vmu xbfc xba feba – jeqvadrakr nerv bx oxda rvmb 100 jvmdmjradu. Eb amjv jeqvadrakr, rvada mda xbfc fxnad farradu mby uqmjau.

Output

Lxd amjv jmua, xirqir xba feba - rva qfmebrakr jxddauqxbyebz rx rva ebqir. Daoaowad rvmr famta uqmjau mfxba.

Sample Input

g g g

Sample Output

q q q

题解：水，但是好难...第一眼看过去，我以为是乱码。。。看看Description最后的网址，vrrq://nnn.uajdarjxyawdamsad.jxo/UJWUxftd.heq，网址应该是http开头的才对，原来这个题目连题目都是经过加密的= =，就从网址入手，得到1/3的密码表，再根据input，output里那些通用的“for each case”“not more than”之类的去找出剩下的密码表，英语不好真心累。

#include <cstdio>

#include <iostream>

#include <string>

#include <sstream>

#include <cstring>

#include <stack>

#include <queue>

#include <algorithm>

#include <cmath>

#include <map>

#define PI acos(-1.0)

#define ms(a) memset(a,0,sizeof(a))

#define msp memset(mp,0,sizeof(mp))

#define msv memset(vis,0,sizeof(vis))

using namespace std;

//#define LOCAL

char a[]=

{

  'e','n','y','r',

  'i','l','q','z',

  'u','c','x','f',

  'a','w','m','j',

  'p','t','k','v',

  's','h','b','o',

  'd','g'

};

int main()

{

#ifdef LOCAL

    freopen("in.txt", "r", stdin);

#endif // LOCAL

    ios::sync_with_stdio(false);

    char c;

    while((c=getchar())!=EOF)

    {

        if(isalpha(c))

        {

            if(isupper(c))

            printf("%c",toupper(a[c-'A']));

            else printf("%c",a[c-'a']);

        }

        else printf("%c",c);

    }

    return ;

}

巴特西

HUST 1358 Uiwurerirexb jeqvad(模拟解密)

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