zoukankan      html  css  js  c++  java
  • pat 1084 Broken Keyboard(20 分)

    1084 Broken Keyboard(20 分)

    On a broken keyboard, some of the keys are worn out. So when you type some sentences, the characters corresponding to those keys will not appear on screen.

    Now given a string that you are supposed to type, and the string that you actually type out, please list those keys which are for sure worn out.

    Input Specification:

    Each input file contains one test case. For each case, the 1st line contains the original string, and the 2nd line contains the typed-out string. Each string contains no more than 80 characters which are either English letters [A-Z] (case insensitive), digital numbers [0-9], or _ (representing the space). It is guaranteed that both strings are non-empty.

    Output Specification:

    For each test case, print in one line the keys that are worn out, in the order of being detected. The English letters must be capitalized. Each worn out key must be printed once only. It is guaranteed that there is at least one worn out key.

    Sample Input:

    7_This_is_a_test
    _hs_s_a_es
    

    Sample Output:

    7TI
     1 #include <iostream>
     2 #include <algorithm>
     3 #include <cstdio>
     4 #include <cstring>
     5 #include <map>
     6 #include <stack>
     7 #include <vector>
     8 #include <queue>
     9 #include <set>
    10 #define LL long long
    11 using namespace std;
    12 const int MAX = 2e3 + 10;
    13 
    14 char s1[MAX], s2[MAX];
    15 int A[MAX] = {0};
    16 
    17 int main()
    18 {
    19 //    freopen("Date1.txt", "r", stdin);
    20     scanf("%s%s", &s1, &s2);
    21     int len1 = strlen(s1), len2 = strlen(s2);
    22 
    23     for (int i = 0; i < len2; ++ i)
    24         if (s2[i] >= 'A' && s2[i] <= 'Z')
    25             A[s2[i]] = 1, A[s2[i] - 'A' + 'a'] = 1;
    26         else if (s2[i] >= 'a' && s2[i] <= 'z')
    27             A[s2[i]] = 1, A[s2[i] - 'a' + 'A'] = 1;
    28         else
    29             A[s2[i]] = 1;
    30 
    31     for (int i = 0; i < len1; ++ i)
    32     {
    33         if (A[s1[i]] == 1) continue;
    34         char a = s1[i];
    35         if (a >= 'a' && a <= 'z')
    36         {
    37             A[a] = 1;
    38             a = char(a - 'a' + 'A');
    39             A[a] = 1;
    40         }
    41         else if (a >= 'A' && a <= 'Z')
    42             A[a] = 1, A[a - 'A' + 'a'] = 1;
    43         else
    44             A[a] = 1;
    45         printf("%c", a);
    46     }
    47     return 0;
    48 }
  • 相关阅读:
    常见的压缩命令
    Linux忘记root密码的解决
    QQFM 中转站(囧转站)OOXX V1.1 by wy811007 (附SkinH_Net的使用) 程序失效 更新1.3版 未发布
    直接插入排序和希尔排序
    SIGABRT错误的调试办法
    UIGestureRecognizer有用文档摘抄
    HTC G14 Sensation Z710e 刷机总结
    iOS 之生命周期(转)
    算法时间复杂度分析基础(转)
    NSURLConnection的同步与异步
  • 原文地址:https://www.cnblogs.com/GetcharZp/p/9591530.html
Copyright © 2011-2022 走看看