CodeForces 408B 手速题

//CodeForces 408B

 1 #include "iostream"
 2 #include "cstdio"
 3 #include "cstring"
 4 #include "algorithm"
 5 using namespace std;
 6 int n1[30], n2[30];
 7 char s1[1010], s2[1010];
 8 int len1, len2;
 9 
10 int main()
11 {
12     int i, j;
13     scanf("%s%s", s1, s2);
14     len1 = strlen(s1);
15     len2 = strlen(s2);
16     for(i = 0; i < len1; ++i) {
17         ++n1[s1[i] - 'a'];
18     }
19     for(i = 0; i < len2; ++i) {
20         ++n2[s2[i] - 'a'];
21     }
22     bool done = 1;
23     int res = 0;
24     for(i = 0; i < 26; ++i) {
25         if(!n1[i] && n2[i]){
26             done = 0;
27             break;
28         }
29         res += min(n1[i], n2[i]);
30     }
31     if(!done)
32         printf("-1
");
33     else
34         printf("%d
", res);
35 }
原文地址:https://www.cnblogs.com/AC-Phoenix/p/4299062.html