谁来帮忙看一下这个题错在哪里？

Winter_moon

Cryptoquote
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Time Limit: 1000 ms Memory Limit: 65536 kB
2 v! w8 N7 \1 x* O# |& Z  L% ^/ nSolved: 79 Tried: 281

# H6 E# |1 b+ @: H, Q8 jDescription
; E' Y0 R1 ]9 }' C  p& rA cryptoquote is a ** encoded message where one letter is simply replaced by another throughout the message. For example:

2 M3 i2 g! J- h! [; l0 cEncoded: HPC PJVYMIY
Decoded: ACM CONTEST
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. K% G. C8 N' a- YIn 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.

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& \5 ^6 K+ p( J) D+ IInput
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.
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4 g, F& ^7 a% A1 nOutput
9 U0 ]2 \1 D( ~/ U7 ^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.

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1 T+ v3 A- q5 Q: H4 oSample Input
2
8 R- o+ a; y$ r- N7 O( d' y% sHPC PJVYMIY
BLMRGJIASOPZEFDCKWYHUNXQTV
3 F$ Z) E. ?- [& |( o, W+ i0 M* S! JFDY GAI BG UKMY
9 [- j9 E; J# d# R9 lKIMHOTSQYRLCUZPAGWJNBVDXEF


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" X' \5 b5 w4 z8 hSample Output
8 S. \; B9 h# [" f0 C1 ACM CONTEST
% ?; I& S0 c0 s9 l) x2 N8 P9 ?  e2 THE SKY IS BLUE

6 v/ n: [2 h: X3 K0 P  {( ?
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+ ?( G# b; w4 G/ g5 t我的代码是：
# r( e0 g' b, V* F: _5 U#include<stdio.h>
#include<string.h>
int main()
{
9 S, C7 [2 M% D% h4 c8 c     int N,i,j,k;
     char c;
     char code[1000];
3 x% O2 ?9 p+ p9 k; S) u+ J     char map[27];
  W, U% R) |6 V0 F+ k4 Q     char trans[1000];
! [3 i4 N0 Q9 E( O2 z; [' ^     char wrong[3];
     scanf("%d",&N);
     gets(wrong);
9 |* V) ^* t3 Y2 q     int h=1;
) Y7 ~/ h# X! F7 e3 l     while(N--)
8 ]; k6 k6 @' M& `' y" a- f$ J     {
         gets(code);
' L$ [9 R* U6 |. u         gets(map);
1 ^! O) ^3 K3 ~# K" B# Y+ @. L* F         for(i=0;((c=code[i])!='\0');i++)
         {
             if(c==' ')trans[i]=' ';
             else {j=code[i]-'A';
             trans[i]=map[j];}       }
/ t2 ?  j  N3 _2 \$ F2 t         printf("%d ",h);
% h" n7 ]1 b% f+ Z+ M$ C+ U         for(i=0;((k=trans[i]<=90)&&(k=trans[i]>=65)||(k=trans[i]==' '));i++)
* e: {& i: w& Z% }. I; `         printf("%c",trans[i]);
1 ?1 Y& d6 f% l6 P/ _         h++;
         printf("\n");
     }
3 b1 t& `( \  v* {     return 0;
! X5 v1 ]# d: H9 C( x}
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" u- z' V7 m1 {样例过了，但是提交答案的时候还是wrong answer。是哪里有陷阱？
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zjqylcy

把答案也附上来吗。。。
& C: @8 N1 @" y4 g3 W6 Y是不是k是int型？

葉_浅浅

.....................

linmatsas

C………………………………

_我还活着

没有看懂.....

huangliao6252

int h=1;这个定义应该在 scanf("%d",&N);前面。