toj 4069 简易统计

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 
 6 const int INF = 1 << 29;
 7 const int N = 10;
 8 const int M = 201;
 9 int cnt1[N];
10 int cnt2[N];
11 char digit[N];
12 char str[M];
13 
14 int solve()
15 {
16     memset( cnt1, 0, sizeof(cnt1) );
17     memset( cnt2, 0, sizeof(cnt2) );
18     for ( int i = 0; i < strlen(digit); i++ )
19     {
20         cnt1[digit[i] - ‘0‘]++;
21     }
22     //将（2，5）视为同一个数字然后计算两次
23     cnt1[2] += cnt1[5], cnt1[5] = cnt1[2];
24     //（6，9）也一样
25     cnt1[6] += cnt1[9], cnt1[9] = cnt1[6];
26     for ( int i = 0; i < strlen(str); i++ )
27     {
28         cnt2[str[i] - ‘0‘]++;
29     }
30     cnt2[2] += cnt2[5], cnt2[5] = cnt2[2];
31     cnt2[6] += cnt2[9], cnt2[9] = cnt2[6];
32     int ans = INF;
33     for ( int i = 0; i < N; i++ )
34     {
35         if ( cnt1[i] == 0 ) continue;
36         int tmp = cnt2[i] / cnt1[i];
37         if ( tmp < ans ) ans = tmp;
38     }
39     return ans;
40 }
41 
42 int main ()
43 {
44     while ( scanf("%s%s", digit, str) != EOF )
45     {
46         int ans = solve();
47         printf("%d\n", ans);
48     }
49     return 0;
50 }