E - I Love You Too解题报告（张宇）

E - I Love You Too
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU
2816

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.

Sample Input

4194418141634192622374
41944181416341926223

Sample Output

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）

[cpp] view
plaincopy

#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++];

else

ans[i]=d[k++];

}

for(i=strlen(a)/2-1;i>=0;i--)

printf("%c",ans[i]);

printf("\n");

}

return 0;

}