Cryptoquote ; k) D, N4 A, j) ^7 |( v" P8 ~/ R
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Time Limit: 1000 ms Memory Limit: 65536 kB( Y. W9 _0 O6 x
Solved: 79 Tried: 281 # m2 _! |5 {+ n H! |! e 1 S5 O4 T: J/ ADescription ) \: q* v) z% C( B; ?; E* H
A cryptoquote is a ** encoded message where one letter is simply replaced by another throughout the message. For example: 9 i4 l0 Y1 a3 N# ?' G . f7 v: e" q% o+ ^1 {( r2 i6 kEncoded: HPC PJVYMIY % x2 }2 G* A4 v+ Z, sDecoded: ACM CONTEST , s+ Z; `8 n4 p+ Y- u4 j/ L( ?' Y' M X" Y. d3 j" h
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. 9 u' C# [/ O) o6 G/ j
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2 F7 D1 _" g( Z, VInput 8 _" _* R+ p2 }5 Y s; {4 m+ u# t+ Y: UThe 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. / b) ~. t: f. T. h/ I8 p8 r+ M5 T2 y6 ^5 I; q" G) x8 W7 I
l8 h/ w9 r& M% l" v# h7 @) o ; P" |( V1 m' x9 l2 ]Output . K% r) A1 P/ ?5 YFor 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. 4 L- u4 M$ k$ m2 F' N) h3 z
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Sample Input ; t x4 h1 b+ h# E" t1 g h W2 ( k6 J1 m/ G# ?' t. M
HPC PJVYMIY 1 ]* `+ V4 a' M I# ]! OBLMRGJIASOPZEFDCKWYHUNXQTV 9 P1 f; g) \2 R- J+ PFDY GAI BG UKMY : ~4 D. M' a" \! XKIMHOTSQYRLCUZPAGWJNBVDXEF( I8 q. n- x# Y
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Sample Output 5 f* F0 M( `! y0 L: F2 ~1 ACM CONTEST d0 t% V5 ]2 g$ S H4 D: k
2 THE SKY IS BLUE( f' O. \! H u+ i9 i
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我的代码是:4 i* v9 z2 `) r* F+ K2 p$ N6 k' Z% q3 k
#include<stdio.h>% j# m$ f f' `! Q _
#include<string.h> & ?' w, r1 L% z9 ]* Rint main() # d Z+ l$ l1 B/ y& O- u7 ?{3 S- v* ?6 D$ }# Q3 i
int N,i,j,k;6 W' [* e9 t0 ? X% p! y! s
char c; - G" W# s2 N# m! B4 \9 a6 f3 d; B char code[1000];) g7 `7 B- J) a% ^3 L' V
char map[27];7 j9 V$ \9 B- ?8 S) i
char trans[1000];$ }6 `" i- Z; X9 W2 K" r
char wrong[3]; 8 O7 Y% C* ?* @$ `8 t3 S- a7 r7 I scanf("%d",&N);# M0 q+ j- U% z- Z
gets(wrong);! @. c6 W* D, d' M& W) Z
int h=1;, o: c$ N; ~( r) E( e
while(N--)6 x: s- |* [; N9 f% Y: |% E0 \. g& k
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gets(code);& J% D2 S* u- a! ?) T3 s4 z+ q
gets(map); 1 V5 [, s2 l, S5 Y$ h for(i=0;((c=code[i])!='\0');i++) 4 H) F; r: U6 W3 `! {6 w {% |" H7 t, s' d) a7 o# z6 I6 x0 n
if(c==' ')trans[i]=' ';( u5 K! V' G9 K2 C: _# n) U
else {j=code[i]-'A';; `9 k7 P, [# W2 A) L
trans[i]=map[j];} } . O6 n- N$ K6 }0 `# R; I printf("%d ",h);! i/ t6 I3 K- v/ O$ J5 C( S, c, n* @
for(i=0;((k=trans[i]<=90)&&(k=trans[i]>=65)||(k=trans[i]==' '));i++)1 K& q% ^( W7 z8 l F% f3 D% K
printf("%c",trans[i]); + M' @5 b! a7 N& [6 r h++;- z+ B7 E j$ _" n' j5 M
printf("\n"); 5 z. s, x6 y% L2 | }- p+ d% r' l, C& n
return 0;+ O' y3 V: C4 }7 b, d. r+ _7 e* h+ X
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