E - I Love You Too解题报告（陈渊）

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
       

 

这个题最麻烦的是数组的初始化，不过还好一次ac。

#include<iostream>
using namespace std;
int main()
{
	char a[10][5],b[1000],c[200],e[1000],f[1000];
	int num,sum,n,m,i,j=0,s,j1;
	a[2][1]='A';a[2][2]='B';a[2][3]='C';a[3][1]='D';a[3][2]='E';a[3][3]='F';
	a[4][1]='G';a[4][2]='H';a[4][3]='I';a[5][1]='J';a[5][2]='K';a[5][3]='L';
	a[6][1]='M';a[6][2]='N';a[6][3]='O';a[7][1]='P';a[7][2]='Q';a[7][3]='R';
	a[7][4]='S';a[8][1]='T';a[8][2]='U';a[8][3]='V';a[9][1]='W';a[9][2]='X';
	a[9][3]='Y';a[9][4]='Z';
	c['Q']='A';c['W']='B';c['E']='C';c['R']='D';c['T']='E';c['Y']='F';
	c['U']='G';c['I']='H';c['O']='I';c['P']='J';c['A']='K';c['S']='L';
	c['D']='M';c['F']='N';c['G']='O';c['H']='P';c['J']='Q';c['K']='R';
	c['L']='S';c['Z']='T';c['X']='U';c['C']='V';c['V']='W';c['B']='X';
	c['N']='Y';c['M']='Z';
	while(cin>>b)
	{
		j=0;
		j1=0;
		s=strlen(b);
		for(i=0;i<s-1;i=i+2)
		{
			j++;
			m=b[i]-'0';
			n=b[i+1]-'0';
			e[j]=c[a[m][n]];
		}
		if(j%2==0)
			num=j/2;
		else
			num=j/2+1;
		m=num+1;
		for(n=1;n<=num;n++)
		{
			f[j1]=e[n];
			j1++;
			if(m<=j)
			{	f[j1]=e[m];
			j1++;
			m++;}
		}
		for(i=j-1;i>=0;i--)
			printf("%c",f[i]);
		cout<<endl;
	}
	return 0;
}