hdu杭电 2816 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.

Sample Input

4194418141634192622374
41944181416341926223

Sample Output

ILOVEYOUTOO
VOYEUOOTIO

//按照题目要求，一步一步模拟，最后输出答案

//做了非常久啊！

<pre name="code" class="cpp">#include<stdio.h>
#include<string.h>
int i,len,j;
char s1[1010],s2[1010];
char s[11][5]={"_","ABC","DEF","GHI","JKL","MNO","PQRS","TUV","WXYZ"};
char str[]={"QWERTYUIOPASDFGHJKLZXCVBNM"};
char s3[1010],s4[1010];
char res[1010];
int main()
{	
	while(gets(s1))
	{
		len=strlen(s1);
		for(i=0,j=0;i<len;i+=2)//数字转换为字母 
		{
			s2[j++]=s[s1[i]-'0'-1][s1[i+1]-'0'-1];
		}
		int k;
		for(i=0;i<j;++i)//按照题目给的方式进行字符变化 
		{
			for(k=0;k<26;++k)
				if(str[k]==s2[i])
				{
					s1[i]='A'+k;break;
				}
		}
		len=len/2+1;
		for(i=0;i<len/2;++i)//把一个字符串分成两 
		{
			s3[i]=s1[i];
		}
		s3[i]='\0';
		int t=i;
		for(j=0;i<len-1;++i)
		{
			s4[j++]=s1[i];
		}
		s4[j]='\0';
		int l1=strlen(s3),l2=strlen(s4);
		
		int flag=1;
		i=0,j=0;
		char ch;
		while(s4[i]!='\0')//轮流存到res数组里 
		{			
			if(!flag)
			{
				flag=1;
				ch=s4[i];
				++i;
				//if(s4[i]=='\0') continue;
			}
			else
			{
				flag=0;
				ch=s3[i];
				
			}
			
			res[j++]=ch;
		}
		if(l1!=l2) res[j++]=s3[i];
		/*
		k=0;
		for(i=0;i<j;++i,k+=2)
		{
			res[k]=s3[i];
			res[k+1]=s4[i];			
		}
		for(;i<t;++i)
		{
			res[k]=s3[i];
			++k;
		}
		*/
		for(i=j-1;i>=0;--i)
		{
			printf("%c",res[i]);
		}
		puts("");
	}
	return 0;
}