hdoj 2816【字符串】

I Love You Too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1807    Accepted Submission(s): 1079

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
   
   
   
   

 

Author

NotOnlySuccess

 

Source

HDU 1st “Old-Vegetable-Birds Cup” Programming Open Contest 

字符串的转化，与字符的位置变换。for循环~

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
int turn(char c)
{
	char yuan[30] = {"QWERTYUIOPASDFGHJKLZXCVBNM"};
	for(int i = 0; i < 26; i++)
		if(c==yuan[i])
		return i+65;
} 
char phone[10][6] = {"ABC","DEF","GHI","JKL","MNO","PQRS","TUV","WXYZ"};
char a[1010],str[506],ser[506];
int main()
{
	int i,j,k,l;
	while(~scanf("%s",a))
	{
		l = strlen(a);
		for(i = 0,k = 0; i < l; i+=2 )
		{
			//printf("%d---%d---\n",a[i]-50,a[i+1]-49); 
			str[k++] = phone[a[i]-50][a[i+1]-49];
		}
		//printf("%d\n",k); 11
		for(i = 0; i < k; i++)
		{
			str[i] = turn(str[i]);
		}
		j = (l/2+1)/2;  
		//printf("%d\n",j); 6
		for(i = j; i < k; i++ )
		{
			ser[i-j] = str[i];
		}
		//str[j] = 0;
		for(i = 0; i < j; i++)
		{
			a[i*2] = str[i];
			a[i*2+1] = ser[i];
		}
		for(i = k-1; i >= 0; i--)
			printf("%c",a[i]);
		printf("\n");
	}
	return 0;
}