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A cryptoquote is a ** encoded message where one letter is simply replaced by another throughout the message. For example:: K2 y: S: \0 j7 E0 f
9 y' A: H( f; o% Q8 {Encoded: HPC PJVYMIY 3 C9 U. ?& v; ]# w$ C
Decoded: ACM CONTEST( I) H1 J9 r1 W: O$ D$ W9 k# p t, C
9 f1 u- g6 H ]' VIn 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. + I" {' y# i) q- y- O# j8 X& j3 `3 o" b6 b$ ?5 ]) h6 f
- Y1 D3 X( e0 q8 L( u" FInput 1 ~9 b( Z9 N# d: `( m0 f4 `4 x1 @
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.' V" M; M6 ?9 A E0 T
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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. : Z: K/ G0 q8 P' ]3 E " i- C% u W* g& i, \/ Z * _; \$ z' r: mSample Input $ ^$ \$ [4 }/ h6 c2 ! D. E8 l X7 y/ AHPC PJVYMIY 8 q2 A, j0 V% K1 U* J! T
BLMRGJIASOPZEFDCKWYHUNXQTV ' l6 ] T9 K* _# }: z% P
FDY GAI BG UKMY 8 @' G3 N1 `0 o' Y& N; ?
KIMHOTSQYRLCUZPAGWJNBVDXEF5 c: J5 x2 G3 J; u4 i N
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Sample Output + ]; F; B1 t( }9 }( F" F- P
1 ACM CONTEST . H& S4 v& C% _( o- [2 THE SKY IS BLUE , Q6 \3 J) f. r. G : t+ L( w' w L- A* w/ h }3 H2 b& x, z% W
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我的代码是： 1 B& N* J) i$ q3 b8 e. j( z#include<stdio.h> * m$ X0 l- `& J+ F# f#include<string.h>5 d Q# _4 G9 a, t
int main() ( k' l/ W* f3 H, E{/ i1 B2 n7 i+ W" \# L, R
int N,i,j,k;) R5 q- h# T' j( j
char c; ' T) I V2 S' G2 t: S n char code[1000]; 1 B+ k; ]: T+ ?* ]- k char map[27]; 4 K G X7 f/ y2 \ char trans[1000];) c$ n" \- o" b- |$ X) H' Q% H& I
char wrong[3]; : G3 s$ d, i! J& v scanf("%d",&N);/ v5 `8 b( H G" w
gets(wrong);3 b6 h3 u3 j6 d" h9 e% b
int h=1; 3 `# m) l# q4 V1 } while(N--)5 _6 V3 P4 l- I, i* x6 r; F! ^
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gets(code); 9 B' c- A7 e" r6 S# j/ s( i* _$ m gets(map); * c' d6 d+ {5 P" H/ I for(i=0;((c=code[i])!='\0');i++)2 p0 @# }8 ~# L' @
{ + v& i1 [3 \, i+ @) Q if(c==' ')trans[i]=' ';$ T& \$ p- \+ k3 U9 e1 I: R
else {j=code[i]-'A';7 ?. f/ a# @" C$ j
trans[i]=map[j];} }; c* T4 w6 I; L6 h/ e9 {
printf("%d ",h); 7 w* L& Z( N) A' T5 Q; j% | for(i=0;((k=trans[i]<=90)&&(k=trans[i]>=65)||(k=trans[i]==' '));i++) & n/ ? }7 k% p3 d- ` printf("%c",trans[i]);. K1 I! R! W& ^1 r0 K
h++; % i* I+ A0 J* u' Z8 V printf("\n");6 G" T ?4 _& e8 t E7 M+ f3 e
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return 0;9 p1 d& {% J% c( K* g
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