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Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 23668 | Accepted: 12945 |
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
CODE:
#include <iostream> #include <cstdio> #include <cstring> #define REP(i, s, n) for(int i = s; i <= n; i ++) #define REP_(i, s, n) for(int i = n; i >= s; i --) #define MAX_N 200 + 10 using namespace std; char s[MAX_N], s1[20]; int l; char plan[27] = {‘0‘, ‘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‘}; int main(){ while(scanf("%s", s1 + 1) != EOF){ if(s1[1] == ‘E‘) break; char s2[20]; getchar(); gets(s + 1); scanf("%s", s2 + 1); l = strlen(s + 1); REP(i, 1, l){ if(s[i] >= ‘A‘ && s[i] <= ‘Z‘) cout << plan[(int)s[i] - ‘A‘ + 1]; else cout << s[i]; } cout << endl; } return 0; }
POJ 1298 The Hardest Problem Ever
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原文地址:http://www.cnblogs.com/ALXPCUN/p/4527397.html