3 p7 X- ^- T/ |5 `6 K/ m , M# n7 o4 R! m" j- RTime Limit: 1000 ms Memory Limit: 65536 kB $ }7 m2 J0 a0 M1 TSolved: 79 Tried: 281 ) E9 n: l3 k C6 H. Z$ v8 X( ^+ F5 |0 g+ D. A( ]/ Z; `4 S' w4 O
Description ' z9 r C- }1 n9 j0 ?9 O# U$ j; ]
A cryptoquote is a ** encoded message where one letter is simply replaced by another throughout the message. For example:3 |# H. g+ M/ U4 e$ c7 d
+ ]- s( q+ b5 j( M7 r; t% D" @Encoded: HPC PJVYMIY ( V H6 x8 G1 r$ y1 m+ N/ y% D
Decoded: ACM CONTEST 1 V6 V2 R; m* e" y2 y. B2 _0 t( x" R 9 S3 o& g/ }) C5 }In 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. & w& s) l! i1 t8 l# W; z
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5 M; F+ C1 h) b7 qInput / X4 S' {, Z8 l: K9 A& c* iThe 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., E+ `" H8 }$ n+ p- @! E
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Output * F, Z# ?: ~% w3 |5 f& m# N
For 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. 7 e* S( q& ?3 Z/ e. f + C5 g$ c2 G9 g1 ^# u. l# k) a! p( N7 \
Sample Input 6 Y& \# R: e5 w1 z( _
2 4 M" F$ ?5 s3 c! _4 A
HPC PJVYMIY 9 v# S- i; d7 E6 ^BLMRGJIASOPZEFDCKWYHUNXQTV 4 f7 b" x n. q( m. eFDY GAI BG UKMY + ?1 p, U! i) ?; ]8 @* f+ Q3 a
KIMHOTSQYRLCUZPAGWJNBVDXEF* S* s2 }, i: B; c/ J
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! v/ e( B2 ]$ @) I5 {' v/ U5 pSample Output + r$ I9 j$ }# U( k" a
1 ACM CONTEST 0 k5 e$ b/ g$ @) C4 Y& g" U9 A
2 THE SKY IS BLUE6 r. b2 J3 a7 U+ l- k
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; O7 T: g k0 Y我的代码是: D( m# D" L: X" w. J#include<stdio.h> ! ~/ B7 H6 v# z, Z1 h$ J+ o#include<string.h> - V" z# Z+ U, T* x9 n# x7 Jint main() 9 D" I7 f5 ?: o3 |6 y" _0 ]{3 y% X# C! }' R9 M$ A2 Z7 Z2 _/ }
int N,i,j,k; $ ^6 j* |: |3 g; W: A2 W char c;( e( E' g% _! P$ x) y9 r' I# o
char code[1000]; 5 C% Z% v! V* P. X+ a* f: U char map[27]; 1 g0 D" |! ~4 X8 m6 @3 k char trans[1000]; ( g2 u8 R/ i4 N, ]6 x char wrong[3];* K9 p" e; r" E3 g' H& [4 P6 O
scanf("%d",&N);0 o% x3 W- k5 K* O6 h1 j! z7 Y
gets(wrong); / W4 R3 t9 |0 { int h=1; ; ~# J# s9 M& U while(N--)% z2 ^1 Q0 p; a9 e
{ & A+ D7 _) t; I, w5 M gets(code); $ k2 w4 n" _! A5 Q6 G gets(map); 6 w+ k) {$ d% Q for(i=0;((c=code[i])!='\0');i++) 7 n Z: q1 z3 Y* q/ [5 m6 R! J {1 _- _$ v& K/ Q$ y" C! a c
if(c==' ')trans[i]=' ';+ R8 q3 ^3 q# a' {1 Z z
else {j=code[i]-'A';8 ?! u5 ]: c2 o# ~/ _
trans[i]=map[j];} } 6 F( U( Y( s" a+ @ printf("%d ",h);+ q1 `! t+ W! j; n3 H* U2 b
for(i=0;((k=trans[i]<=90)&&(k=trans[i]>=65)||(k=trans[i]==' '));i++) * m, Y! V- F6 M9 R7 H! j printf("%c",trans[i]);+ i- c5 A5 [ o; T" ]+ G
h++; 4 D& x7 X7 C" |5 X1 {; f printf("\n");( `& }# [6 H4 u9 x8 b
} 9 U" k" z U3 b+ V" o: [: Q7 ~! P return 0;3 w% F' V" q+ \+ T
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