Cryptoquote 7 l: n9 \1 L) L5 C; q9 W0 N. O& d" @2 D* G% a
( z; Y: Q2 T/ e6 gTime Limit: 1000 ms Memory Limit: 65536 kB ; S/ i: a2 y! k! s B1 nSolved: 79 Tried: 281 $ j2 k7 S& {( z: g2 D, z& A+ Y3 D! L! u0 t A) l! C' j
Description 7 \" B, g) R. I$ W
A cryptoquote is a ** encoded message where one letter is simply replaced by another throughout the message. For example:! F* t3 Z$ B" l/ B2 E
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Encoded: HPC PJVYMIY , A' D/ K) Y- o9 G2 gDecoded: ACM CONTEST . n0 Z. J4 K# R; U" ] e4 p, p6 [5 B3 y% b! h. NIn 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. . I1 W. X0 R+ h6 C6 e( Z* H+ ^( S. r/ u7 Q3 D
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Input 0 T! g8 H0 E9 u! N8 {
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. 9 ]; A* i# T9 ~9 p# k & q* U5 v3 A8 s5 _8 E8 _+ j& F$ R; k% r3 @, G
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Output - Z/ [! a& C ], T/ C* f$ Z: s
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. . {0 @2 r2 B: O/ Y& k0 i8 B$ v
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Sample Input & `9 P) q; z, x8 H
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HPC PJVYMIY ( x7 K: x$ N6 a2 [BLMRGJIASOPZEFDCKWYHUNXQTV ( }# K; {: e# x+ w# ^6 c$ G6 |FDY GAI BG UKMY $ j7 h4 `6 o6 a' T# N3 OKIMHOTSQYRLCUZPAGWJNBVDXEF/ V6 q# i, s) u" L% u
# f" x) i9 }6 m; v # d/ D' h7 i7 X : x! Y* J4 F* }1 P/ oSample Output 3 R' S% k2 O& W1 ACM CONTEST , z7 S5 Z p. E
2 THE SKY IS BLUE 9 s6 m j, G( } 4 @5 s3 `- X y& A- ~ # S+ U8 m# e" c0 A7 k3 ~' y! p$ D5 K7 l; Y2 N& {) w2 H; ~ ]
我的代码是:7 z7 [5 ?2 ]. s! L/ j) |
#include<stdio.h>7 b9 p( k6 Y% f q; Z1 H6 a' b
#include<string.h>, r5 Z7 P% p, U
int main()4 v6 C$ p: T4 g+ |4 X
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int N,i,j,k; 2 d8 n5 A" A9 r | char c; , k0 V) ^1 E0 J$ a! ~ char code[1000]; * K) j+ y+ x3 [5 A2 s; H char map[27];* L0 E9 Y V5 P, e
char trans[1000]; ) o8 A" b" g" H3 |: L2 A char wrong[3]; 9 w5 }: F6 }! d0 p0 O: D! f scanf("%d",&N); 3 H! O- X5 K$ }6 b gets(wrong);0 M" F/ ?# h1 a% q' L
int h=1; # J L; \0 Y9 n$ D5 z" T9 x while(N--) + u' {( t8 W3 V- K { , A% L, T% x: J" f# x5 L7 z" I gets(code); : O7 W) i* @6 K8 ?, b1 |: c gets(map); 5 [. }# ]" @$ r$ Q. n3 z for(i=0;((c=code[i])!='\0');i++)1 ?" i4 s! \, o* Z
{ 7 u( e% S+ z) T: o8 G, ?$ @8 j% ~ if(c==' ')trans[i]=' '; 4 }5 b" q. r- X else {j=code[i]-'A';' D) J( B9 U5 H; Q3 L9 D( F5 z4 U
trans[i]=map[j];} } - M4 h6 X4 a B$ K8 Y printf("%d ",h);% U5 @0 S v6 q- |5 T
for(i=0;((k=trans[i]<=90)&&(k=trans[i]>=65)||(k=trans[i]==' '));i++)' b5 R2 U) \/ I
printf("%c",trans[i]); ! {/ i7 H( P7 b& V) T' t h++; w) P; t. H$ Z* x) Y
printf("\n");7 I% m/ H* j( i7 P
} ) x) k4 s- }" d return 0; . Z+ t5 _ q( z) x: y}! G4 U( d* l- G; p" r