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Time Limit: 1000 ms Memory Limit: 65536 kB. \% H4 z- h8 ^
Solved: 79 Tried: 281 3 T- {8 N3 `5 b + B! l4 x1 A& h3 t' qDescription , D+ j) P. N/ I$ f- V6 q0 p& O
A cryptoquote is a ** encoded message where one letter is simply replaced by another throughout the message. For example: ; C$ o5 m9 [; o' n+ }/ ?6 P3 C5 }9 S 3 l+ }. `. P6 n9 V6 JEncoded: HPC PJVYMIY & A7 L: H- h, A# I
Decoded: ACM CONTEST # `0 x1 k, n1 r. x- B7 I" O6 P 5 K( R- z" N2 t6 SIn the example above, H=A, P=C, C=M, J=O, V=N, Y=T, M=E and I=S. For this problem, you will decode messages. 6 t8 H3 f" K( ?3 w6 B9 U( A + o6 L1 g$ x/ n! ^1 \; E . r1 T4 a" ^/ W- [9 v! I8 EInput + x6 q) u' W' [The first line of input contains a single integer N,(1<=N<=1000) which is the number of data sets that follow. Each data set consists of two lines of input. The first line is the encoded message. The second line is a 26 character string of upper case letters giving the character mapping for each letter of the alphabet: the first character gives the mapping for A, the second for B and so on. Only upper case letters will be used. Spaces may appear in the encoded message, and should be preserved in the output string. I* f2 o) P/ h4 J: D- P' V9 w- {7 M( Q* L) d, |
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Output $ ]2 ?+ g$ D( p+ D* iFor each data set, you should generate one line of output with the following values: The data set number as a decimal integer (start counting at one), a space and the decoded message. $ c, |, L! A& R, F: O% E
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Sample Input 4 W! ~* _& i+ x* W, n& H2 . X, m \- K6 l: }" H
HPC PJVYMIY M5 K9 g/ e! t, Z1 xBLMRGJIASOPZEFDCKWYHUNXQTV * V, E3 P8 I& h6 P) |3 L( x) b/ a1 V
FDY GAI BG UKMY 2 m: C( A6 r3 E+ a& L- N; ^KIMHOTSQYRLCUZPAGWJNBVDXEF 9 U& a) A3 S; n) m. P0 K( D5 n1 G' Y6 @+ [. d. J8 o
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Sample Output 1 J8 _' b' ~* F5 u. V& f4 G" g1 ACM CONTEST 6 t( A$ n: d; A7 H( j, o
2 THE SKY IS BLUE * G$ I) j$ ]3 {) o / t! ^% {- J1 k$ K3 r/ i/ k. a# i) a# ]* ~4 `( M
9 l4 z% t7 l8 E, v我的代码是:7 Z' [) z, O; R$ {
#include<stdio.h>+ ?. R5 C, {5 Q" k7 ^& Y5 p
#include<string.h>8 H! u% t" G( f& X' A; v E
int main() 8 l5 H/ C- t# m B+ N h{; Y7 O1 o4 p- D
int N,i,j,k; 3 R. M2 z/ B8 F6 o" K. ?4 J char c; ' ]! U$ U+ i w& W char code[1000];; f" C( R* t. R: Z9 r8 w
char map[27];, b; ]$ J$ D& b
char trans[1000]; ; K: W9 ~: Y# u. O, s. O char wrong[3];& R2 ~5 p- s8 ]* M# R r+ w" v
scanf("%d",&N);6 B- P9 K, k: B$ }7 E) h6 ?
gets(wrong); * I% p/ M# U7 V+ G; X int h=1; 3 S/ H( T7 s- B0 m* b while(N--) % O0 `1 U+ J7 X: ~( O( Q+ K { 0 U+ j; ]3 K3 t gets(code);" [- D3 W) w' O
gets(map); ' w+ n6 c4 V& E' E* C, C for(i=0;((c=code[i])!='\0');i++)2 y6 ]5 Q* Z8 N. p: r" P
{ % `0 @$ h' }) C1 \! k, ^: o: U& i if(c==' ')trans[i]=' ';4 ?' Q* j6 G- O# V% b# d; H4 o
else {j=code[i]-'A';* [& p8 X4 R& Z' l% \- g* ?* y& ^1 m
trans[i]=map[j];} }$ U1 ]% A' n* u" r! U* s' G
printf("%d ",h);# Y! N0 K8 [1 Y8 ?+ } C
for(i=0;((k=trans[i]<=90)&&(k=trans[i]>=65)||(k=trans[i]==' '));i++) 7 p9 r: p, ?8 y* E; Z) w0 q printf("%c",trans[i]); ; o) X9 _. G, Z2 C% K h++; & m5 y# d: K# ]4 E& K printf("\n"); 1 i: m) W( O9 i: E } " `2 f# D: a7 f+ _ return 0;% n) X/ R- p! ?% P. u0 m3 r
} - Y, L% f9 f: F) ]( Y8 D 9 D0 F5 a( h- t# | ! z' E4 ], r" a3 s; _. g; u样例过了,但是提交答案的时候还是wrong answer。是哪里有陷阱? 0 [# v( L9 \% _; O5 m