标题: 谁来帮忙看一下这个题错在哪里？ [打印本页] 作者: Winter_moon 时间: 2011-3-19 15:26 标题: 谁来帮忙看一下这个题错在哪里？ Cryptoquote 4 {' S7 V/ e7 [3 O o $ r, E# c% z, b5 n# [0 n8 L, R" O" ~% Y* H6 f( g
Time Limit: 1000 ms Memory Limit: 65536 kB) l. L4 w; A, x' h# E) e
Solved: 79 Tried: 281 $ i0 { n4 r/ i; `( n2 N- n0 [6 \/ Q9 d7 ^$ \) H( v- r: U8 p
Description $ ]. @# r% {+ Y: T- NA cryptoquote is a ** encoded message where one letter is simply replaced by another throughout the message. For example:! @# ?" k5 e5 k& \
! J6 }& F: W4 {2 \ NEncoded: HPC PJVYMIY & w, C8 {2 D+ @* ^
Decoded: ACM CONTEST , e2 j9 |4 l+ j" m$ o) Q3 Q! @. q( J Z6 V, H) x6 {0 N
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. 6 j) m5 K) j. b8 [# Q$ f0 n 0 a3 r+ X3 W1 t1 t , `2 }- S. H0 l3 kInput " A7 X& U8 i- f, r5 c0 Q! I# X4 f8 Z
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. : W& P1 L5 X- z2 ~3 _3 U& @8 Q5 v5 I
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$ @2 A7 K9 I, ]Output ( v' Q" U0 V3 JFor 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 b! g. S. n( s- q" @
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Sample Input " b7 Z, q# u# X# O: L% m
2 4 Q: ~! H* H" _: C; ]% zHPC PJVYMIY , f/ M) A6 C' `. E g6 R7 G
BLMRGJIASOPZEFDCKWYHUNXQTV 0 s4 _+ M1 z' u9 z& Y+ _
FDY GAI BG UKMY ; M/ ?% g4 e2 H' g* k* ^
KIMHOTSQYRLCUZPAGWJNBVDXEF + z- ]8 k# U, {" [0 I 9 X+ S$ w6 l& d u8 k1 u $ X7 e: @# p! G& V$ m 7 E U+ R' ^4 P1 `4 Q GSample Output 7 F& M. k& u+ e1 q3 a- f
1 ACM CONTEST * s! j3 \* X9 Y) Z9 T0 n
2 THE SKY IS BLUE" p* N) W" J6 [1 z% R& _ h; m
0 ^) c0 H( E! t2 d5 Z- c2 D
/ A. \) h0 l0 \$ O+ z( ~ 0 E- `. J$ v) K* n( Y& r! ~我的代码是： ( x2 X( C8 K% ~( `" ^9 k#include<stdio.h> ' H1 z- m0 G; Z- {( J/ B( W#include<string.h>6 D2 T! b& u6 @% c
int main()9 o2 }5 H8 E# k6 a& Z9 R
{ 5 H7 g0 U% N9 w' @7 K; ]$ _ int N,i,j,k; - i. y, A0 G# I( o5 ]# Y A3 W' a* I char c;! q" y! Y+ z7 ~0 U# F7 h9 U
char code[1000];8 z& s: @% D6 _+ P! S8 a
char map[27]; " m% E3 C5 {( T9 y0 @6 o$ o char trans[1000]; * K: V- q) q/ T# o& y( K2 u) p0 A char wrong[3];; ~4 h1 t! b6 C
scanf("%d",&N);' z8 |' [: I# w6 u, J3 `
gets(wrong); - D$ x! \1 S: s( d8 e int h=1; # D; {9 o6 D( S. x; r while(N--) ( W; G# x+ F# I$ | {, L) h' d. z9 z5 k1 l' P
gets(code); ; D! j; g- Q/ R- x gets(map); 8 V& C; D, r1 p4 l for(i=0;((c=code[i])!='\0');i++)" M) N; ?# ?) r( C3 ]8 J
{ + X1 M8 e* Q$ R- D( Y. w) r if(c==' ')trans[i]=' '; m) B j9 `+ f9 s1 a% Z% S) C else {j=code[i]-'A';3 c# k2 M6 v. N- F! h2 S: Z% V* {$ L8 z
trans[i]=map[j];} }. g- `7 `& J1 T) N9 P# R2 G5 O
printf("%d ",h);; x* d; L, s7 n# w$ @+ T! g: X
for(i=0;((k=trans[i]<=90)&&(k=trans[i]>=65)||(k=trans[i]==' '));i++)( d3 i1 O5 B; Y0 p% n& \
printf("%c",trans[i]);7 }6 m3 C0 Y* V3 l
h++; # `8 @8 Z- u+ u- | printf("\n");3 d# K Y: Z, Y& X
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return 0;! ?7 S9 j5 h" }0 g: Z
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