杭电ACM HDU 2816 I Love You Too
2013-06-19 22:26
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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"); } }
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