hdu-2816 I Love You Too

                            I Love You Too

Problem Description

This is a true story. A man showed his love to a girl,but the girl didn't replied clearly ,just gave him a Morse Code:
****-/*----/----*/****-/****-/*----/---**/*----/****-/*----/-****/***--/****-/*----/----*/**---/-****/**---/**---/***--/--***/****-/   He was so anxious that he asked for help in the Internet and after one day a girl named "Pianyi angel" found the secret of this code. She translate this code as this five steps:
1.First translate the morse code to a number string:4194418141634192622374
2.Second she cut two number as one group 41 94 41 81 41 63 41 92 62 23 74,according to standard Mobile phone can get this alphabet:GZGTGOGXNCS



3.Third she change this alphabet according to the keyboard:QWERTYUIOPASDFGHJKLZXCVBNM = ABCDEFGHIJKLMNOPQRSTUVWXYZ
So ,we can get OTOEOIOUYVL
4.Fourth, divide this alphabet to two parts: OTOEOI and OUYVL, compose again.we will get OOTUOYEVOLI
5.Finally,reverse this alphabet the answer will appear : I LOVE YOU TOO

I guess you might worship Pianyi angel as me,so let's Orz her.
Now,the task is translate the number strings.

Input

A number string each line(length <= 1000). I ensure all input are legal.

Output

An upper alphabet string.

 

SampleInput

4194418141634192622374

41944181416341926223

SampleOutput

ILOVEYOUTOO

VOYEUOOTIO

题意：给出一串长度不大于1000的数字，按每两个数字对应一个字母来解密（如图，即手机按键上的字母，如22代表B），解密后在根据解密对应的编码【3】来得到字母串，然后拆成两半（前部分不短于后部分），再按照【4】重新得到字母串，最后反序输出该字母串就ok啦。（貌似此密码很让人头疼呀。。。。别急，一步一步来，题目并不难）

#include<iostream>using namespace std;int main(){char a[1010];char b[505];char c[252];char d[252];char ans[505];char x[8][5]={"KXV","MCN","OPH","QRS","ZYI","JADL","EGW","BUFT"}; //已经按照密码定义了while(scanf("%s",a)!=EOF){int i,j,k;for(i=0,j=0;i<strlen(a);i+=2,j++){a[i]-='0';            //把字符转化为数字a[i+1]-='0';b[j]=x[a[i]-2][a[i+1]-1];   //求出每两个数字对应的一个字母}//要把字母串分成两个部分,而且前部分不短于后部分for(i=0;i<((strlen(a)/2)+1)/2;i++)c[i]=b[i];for(j=i,k=0;j<strlen(a)/2;j++,k++)d[k]=b[j];j=0;k=0;//再把两个字母串合并for(i=0;i<strlen(a)/2;i++){if((i+1)%2!=0)ans[i]=c[j++];elseans[i]=d[k++];}for(i=strlen(a)/2-1;i>=0;i--)printf("%c",ans[i]);printf("\n");}return 0;}