$ _; H3 \" j; y5 ^# E: d/ \( VTime Limit: 1000 ms Memory Limit: 65536 kB : Z2 d. Y T/ i7 Y4 e9 d. A% WSolved: 79 Tried: 281 8 K0 s* ] t; n6 P
4 S) r5 a7 G- E1 S7 b+ G1 q# JDescription , W% v! _0 s1 n) b) Q* P
A cryptoquote is a ** encoded message where one letter is simply replaced by another throughout the message. For example: / L- ~/ f9 q/ x5 W2 M9 e" [# K8 E' ~ 6 e3 |( Q7 @) o, d: ?Encoded: HPC PJVYMIY 4 I7 P2 {. U+ |3 @: E4 VDecoded: ACM CONTEST / }' |9 Q& X) [) n4 X* b ! k. P5 q. V; @5 UIn 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. / a$ i+ U. T: t1 c% C5 ~ 9 S' v. d) Y: b! F6 m. a0 k6 \" G . P; @& M9 Q, J9 | eInput ; q2 P2 n* @ 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. 0 [2 z6 k9 t5 b0 m# v5 I" N# F) D9 H, |5 q/ _9 S( F
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Output - [/ y- H* x2 [. {# T: hFor 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. # ]6 E; E' e2 ^, E% y+ D 6 q! q H9 Z" V ( m3 H2 z: U( o: P: HSample Input . L+ t: [1 |! O" ?! \. c2 ! a8 \2 |; }' K' lHPC PJVYMIY 4 x/ Z* s ? }' b) M- I% nBLMRGJIASOPZEFDCKWYHUNXQTV : v8 @5 E* A1 e$ p( M _, K) ?
FDY GAI BG UKMY & ^" G: @' W9 hKIMHOTSQYRLCUZPAGWJNBVDXEF Z1 M, X; z! D8 b9 o1 e% N* g- d/ R6 N X; d; n. R$ N5 S( V3 h
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Sample Output 8 l: n; _3 ~9 v5 _, d1 L/ E/ N
1 ACM CONTEST 7 E! v: {0 Q+ l7 V1 N3 l. n
2 THE SKY IS BLUE4 u: e Z) a9 [8 o( }+ C
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我的代码是：' m% R3 i2 u9 d$ ~1 s% \4 d
#include<stdio.h> / O! i# U: K& ^#include<string.h> " |% {/ y0 U( B) Y/ Y0 U$ U+ C2 Mint main() : ~& D$ H/ r5 [# ?1 z# w# w& m% N{ v( m, k/ L% P8 B int N,i,j,k; 7 k6 i% F5 u, @# E# D char c; / {3 ~# N0 [& U char code[1000]; 7 I6 l5 b I6 B+ x8 W) e char map[27];# A7 M, E L! T3 f- y( u
char trans[1000];$ Q7 {$ Y1 T" ^
char wrong[3]; ) a( E% w; F K9 b. ` scanf("%d",&N); $ y3 B `1 c+ f o( c9 F gets(wrong); 3 M7 [! G- W. c$ s! }8 i- | int h=1; 6 e; t; b# y' A while(N--) 2 A/ L. }+ c# u D& M5 j3 e {$ u" D$ y7 o* s
gets(code); ' B' I/ s" v: e7 X; \4 l* m4 @8 N gets(map);# ~* a" V. Y1 @3 s
for(i=0;((c=code[i])!='\0');i++) / b: Y8 ?/ x+ R7 i9 d {3 D0 _) |& \/ j! q
if(c==' ')trans[i]=' '; % S- |8 \% P" b0 A+ @+ _ else {j=code[i]-'A'; ! Z( N( V$ u% y7 ~* d' y trans[i]=map[j];} } 7 J4 J0 G% _. R printf("%d ",h); / C: Z3 h$ L( G. a3 L for(i=0;((k=trans[i]<=90)&&(k=trans[i]>=65)||(k=trans[i]==' '));i++) ( [1 S- g# U! }+ c v: x$ m printf("%c",trans[i]);4 U4 T* L+ W {/ ^1 X- _/ ~" d6 E
h++;0 j Q! M( Q. O2 U6 t% @4 r3 c
printf("\n");- {( b, I/ \7 j+ @ X* M' a
} % R1 y# q1 m0 ~* W1 W Z return 0; * A% O4 j: K3 N2 R}& q# D+ ?/ L1 \# s6 |6 t' o