查看: 3834|回复: 5

谁来帮忙看一下这个题错在哪里？

字体大小: 正常放大

1 主题	0 听众	2 积分

升级 40%

该用户从未签到

电梯直达

1^#

发表于 2011-3-19 15:26 |只看该作者 |倒序浏览

|招呼Ta 关注Ta

Cryptoquote

Time Limit: 1000 ms Memory Limit: 65536 kB
Solved: 79 Tried: 281

Description
A cryptoquote is a ** encoded message where one letter is simply replaced by another throughout the message. For example:

Encoded: HPC PJVYMIY
Decoded: ACM CONTEST

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.

Input
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.

Output
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.

Sample Input
2
HPC PJVYMIY
BLMRGJIASOPZEFDCKWYHUNXQTV
FDY GAI BG UKMY
KIMHOTSQYRLCUZPAGWJNBVDXEF

Sample Output
1 ACM CONTEST
2 THE SKY IS BLUE

我的代码是：
#include<stdio.h>
#include<string.h>
int main()
{
   int N,i,j,k;
   char c;
   char code[1000];
   char map[27];
   char trans[1000];
   char wrong[3];
   scanf("%d",&N);
   gets(wrong);
   int h=1;
   while(N--)
   {
      gets(code);
      gets(map);
      for(i=0;((c=code[i])!='\0');i++)
      {
         if(c==' ')trans[i]=' ';
         else {j=code[i]-'A';
         trans[i]=map[j];}    }
      printf("%d ",h);
      for(i=0;((k=trans[i]<=90)&&(k=trans[i]>=65)||(k=trans[i]==' '));i++)
      printf("%c",trans[i]);
      h++;
      printf("\n");
   }
   return 0;
}

样例过了，但是提交答案的时候还是wrong answer。是哪里有陷阱？

zan