(I2
I57
I5
(dp0
Vnecklace.%l
p1
(Vnecklace.cpp
p2
S'#include <iostream>\r\n#include <cstdio>\r\nusing namespace std;\r\nlong long n, k, letters[100000][27], counter = 0;\r\n\r\nint main()\r\n{\r\n    string s, s1;\r\n    pair <char, char> p[677];\r\n\r\n    scanf("%lld %lld", &n, &k);\r\n    cin >> s;\r\n    //scanf("%s\\n", &s);\r\n    //cout << s << endl;\r\n\r\n    for(int i = 0; i < k; ++i)\r\n    {\r\n        cin >> s1;\r\n\r\n        p[i].first = s1[0];\r\n        p[i].second = s1[1];\r\n    }\r\n\r\n    for(int j = 0; j < 26; ++j)\r\n    {\r\n        letters[0][j] = 0;\r\n    }\r\n\r\n    letters[0][s[0] - \'a\']++;\r\n\r\n    for(int i = 1; i < s.size(); ++i)\r\n    {\r\n        for(int j = 0; j < 26; ++j)\r\n        {\r\n            letters[i][j] = letters[i - 1][j];\r\n        }\r\n\r\n        letters[i][s[i] - \'a\']++;\r\n    }\r\n\r\n    for(int i = 0; i < k; i++)\r\n    {\r\n        for(int j = 0; j < s.size(); j++)\r\n        {\r\n            if(s[j] == p[i].first)counter += (letters[s.size() - 1][p[i].second - \'a\'] - letters[j][p[i].second - \'a\']);\r\n        }\r\n    }\r\n\r\n    printf("%lld\\n", counter);\r\n\r\n\r\n    return 0;\r\n\r\n}\r\n'
p3
tp4
stp5
.