HDU 4690 EBCDIC 2013 Multi-University Training Contest 9

解题报告：一个模拟题，有两张表格，然后输入一个字符在第一章表格中的位置，让你找出这个字符在第二章表对应的位置。

我欧诺个的是暴力打表，输了两百多个数字，时间复杂度直接降到O（1)，这题觉得比较坑的就是题目里面没有给出数据范围，一开始数组开小了，一直WA，搞得我一直以为是表打错了，找了半个小时，结果确定正确，才怀疑是数组开小了。这题数组至少要开到10的七次方。

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 using namespace std;

 int map[][] = {
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 {,,,,,,,,,,,,,,,},
 };

 char temp[] = {'','','','','','','','','','','A','B','C','D','E','F'};
 char str[];
 int main()
 {
     while(scanf("%s",str)!=EOF)
     {
         int len = strlen(str);
         for(int i = ;i<len;i+=) {
             int x,y;
             if(str[i] >= '' && str[i] <= '')
             x = str[i] - '';
             else x = str[i] - 'A'+;
             if(str[i+] >= '' && str[i+] <= '')
             y = str[i+] -'';
             else y = str[i+] - 'A'+;
             int d = map[x][y];
             int m = d / ;
             int n = d % ;
             printf("%c%c",temp[m],temp[n]);
         }
         puts("");
     }
     return ;
 }