UVALive 4174

DES:给出一个字符串。连续空格的个数代表一个新的字符。奇数个表示0。偶数个表示1。然后根据这个码作为ASCII码。写出对应的字符。就是统计空格个数。二进制转换成十进制的小模拟。但是比赛的时候敲得很不顺。

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<map>
  4 #include<vector>
  5 #include<math.h>
  6 #include<iostream>
  7 #include<map>
  8 using namespace std;
  9
 10 char str[100];
 11 int num[400];
 12 char anss[70];
 13 int cou;
 14 int tt;
 15
 16 void change()
 17 {
 18     int temp = (cou % 5);
 19     int ans = 0;
 20     tt = 0;
 21     char la;
 22     for (int i=0; i<temp; ++i)
 23     {
 24         num[cou++] = 0;
 25     }
 26     for (int i=0; i<cou; i+=5)
 27     {
 28         int t = 4;
 29         ans = 0;
 30         for (int j=i; j<i+5; ++j)
 31         {
 32             ans += pow(2, t) * num[j];
 33             t--;
 34         }
 35         if (ans >=1 && ans <= 26)
 36         {
 37             int hh = ans + int(‘A‘);
 38             hh -= 1;
 39             la = char(hh);
 40         }
 41         else if (ans == 30) la = ‘.‘;
 42         else if (ans == 29) la = ‘-‘;
 43         else if (ans == 28) la = ‘,‘;
 44         else if (ans == 27) la = ‘\‘‘;
 45         else if (ans == 0) la = ‘ ‘;
 46         else if (ans == 31) la = ‘?‘;
 47         anss[tt++] = la;
 48     }
 49 }
 50
 51
 52 int main()
 53 {
 54     cou = 0;
 55     while(gets(str))
 56     {
 57         int len = strlen(str);
 58         int cnt = 0;
 59         if (str[0] == ‘*‘)
 60         {
 61             change();
 62             int st = 0;
 63             while(anss[st] == ‘ ‘)
 64                 st++;
 65             int ed = tt-1;
 66             while(anss[ed] == ‘ ‘)
 67                 ed--;
 68             for (int i=st; i<=ed; ++i)
 69             {
 70                 cout << anss[i];
 71             }
 72             cout << endl;
 73             memset(num, 0, sizeof(num));
 74             memset(anss, 0, sizeof(anss));
 75             cou = 0;
 76             continue;
 77         }
 78         if (str[0] == ‘#‘)
 79             break;
 80
 81         for (int i=0; i<len; ++i)
 82         {
 83             cnt = 0;
 84             if (str[i] == ‘ ‘)
 85             {
 86                 int j;
 87                 for (j=i; ; j++)
 88                 {
 89                     if (str[j] != ‘ ‘)
 90                         break;
 91                     cnt++;
 92                 }
 93                 if (cnt % 2 == 1) num[cou++] = 0;
 94                 else num[cou++] = 1;
 95                 cnt = 0;
 96                 i = j;
 97             }
 98         }
 99     }
100     return 0;
101 }

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