杭电ACM HDU 2816 I Love You Too

I Love You Too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1258    Accepted Submission(s): 759

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

Recommend

lcy

因为不仔细，WA了1次

#include<cstdio>
#include<cstring>
int turn(char c){
    int i;
    char key[30]={"QWERTYUIOPASDFGHJKLZXCVBNM"};
    for(i=0;i<26;i++)
        if(c==key[i])return i+65;
}
int main(){
    int a,i,l;
    char p[10][5]={"ABC","DEF","GHI","JKL","MNO","PQRS","TUV","WXYZ"};
    char num[1001],str1[501],str2[501];
    while(~scanf("%s",num)){
        l=strlen(num);
        for(i=0;i<l;i+=2){
            str1[i/2]=p[num[i]-50][num[i+1]-49];
        }
        for(i=0;i<l/2;i++)
            str1[i]=turn(str1[i]);
        for(i=(l/2+1)/2;i<=l/2;i++)
            str2[i-(l/2+1)/2]=str1[i];
        str1[(l/2+1)/2]=0;
        for(i=0;i<l/4+1;i++){
            num[i*2]=str1[i];
            num[i*2+1]=str2[i];
        }
        for(i=l/2-1;i>=0;i--)
            printf("%c",num[i]);
        printf("\n");
    }
}