The Hardest Problem Ever
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14589 Accepted Submission(s): 6709
You are a sub captain of Caesar's army. It is your job to decipher the messages sent by Caesar and provide to your general. The code is simple. For each letter in a plaintext message, you shift it five places to the right to create the secure message (i.e., if the letter is 'A', the cipher text would be 'F'). Since you are creating plain text out of Caesar's messages, you will do the opposite:
Cipher text
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Plain text
V W X Y Z A B C D E F G H I J K L M N O P Q R S T U
Only letters are shifted in this cipher. Any non-alphabetical character should remain the same, and all alphabetical characters will be upper case.
A single data set has 3 components:
Start line - A single line, "START"
Cipher message - A single line containing from one to two hundred characters, inclusive, comprising a single message from Caesar
End line - A single line, "END"
Following the final data set will be a single line, "ENDOFINPUT".
START NS BFW, JAJSYX TK NRUTWYFSHJ FWJ YMJ WJXZQY TK YWNANFQ HFZXJX END START N BTZQI WFYMJW GJ KNWXY NS F QNYYQJ NGJWNFS ANQQFLJ YMFS XJHTSI NS WTRJ END START IFSLJW PSTBX KZQQ BJQQ YMFY HFJXFW NX RTWJ IFSLJWTZX YMFS MJ END ENDOFINPUT
IN WAR, EVENTS OF IMPORTANCE ARE THE RESULT OF TRIVIAL CAUSES I WOULD RATHER BE FIRST IN A LITTLE IBERIAN VILLAGE THAN SECOND IN ROME
DANGER KNOWS FULL WELL THAT CAESAR IS MORE DANGEROUS THAN HE
#include<stdio.h> #include<string.h> int main() { char s1[1000]; int i,len; int t; while(true) { gets(s1);//获取文本 // printf("%s",s1); if(strcmp(s1,"ENDOFINPUT")==0)//结束控制 break; if(strcmp(s1,"START") != 0 && strcmp(s1,"END") !=0)//获取密文 { len = strlen(s1);//获取长度 for(i = 0;i < len;i++) { // printf("%d",s1[i]); if(s1[i] >= 'A' && s1[i] <= 'Z') { t = (s1[i]-'A'+21) % 26; printf("%c",t+'A'); } else printf("%c",s1[i]); } printf("\n"); } } return 0; }