CF 1379 A. Acacius and String

Posted 2020-12-18 ssummerzzz

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题目：可以改变‘?‘为任意‘a‘~‘z‘的字符，可不可以让s有且仅有一个子串为"abacaba"。

思路：暴力就行，枚举每个位置开始7个字符能否组成"abacaba"，可以的话在判断此时把这7个位置的字符变成"abacaba"时，s有几个"abacaba"子串。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <queue>
 5 #include <string>
 6 #include <vector>
 7 #include <cmath>
 8  
 9 using namespace std;
10  
11 #define ll long long
12 #define pb push_back
13 #define fi first
14 #define se second
15 
16 const int N = 2e5 + 10;
17 char s[N];
18 int len;
19 
20 void solve()
21 {      
22     string p = "abacaba";
23     int T;
24     cin >> T;
25     while(T--){
26         string s, tmp;
27         int n;
28         cin >> n >> s;
29 
30         int good = 0;
31         for(int i = 0; i < n; ++i){
32             if(i + 6 >= n) break;
33 
34             //几个字符匹配
35             int ok = 0;
36             for(int j = 0; j < 7; ++j){
37                 if(s[i + j] == ‘?‘ || p[j] == s[i + j]) ok++;
38                 else break;
39             }
40             if(ok == 7){
41                 tmp = s;
42                 for(int j = 0; j < 7; ++j) tmp[i + j] = p[j];
43 
44                 //出现几次    
45                 int cnt = 0;
46                 string::size_type inx = 0;
47                 while(1){
48                     inx = tmp.find(p, inx);
49                     //printf("inx = %d
", (int)inx);
50                     if(inx == tmp.npos) break;
51                     inx = inx + 3;
52                     cnt++;
53                 }
54                 if(cnt == 1){
55                     good = 1;
56                     cout << "yes" << endl;
57                     for(int j = 0; j < n; ++j){
58                         cout << (tmp[j] == ‘?‘ ? ‘z‘ : tmp[j]);
59                     }
60                     cout << endl;
61                     break;
62                 }
63             }
64         }
65 
66         if(!good) cout << "no" << endl;
67     }
68 }
69  
70 int main()
71 {
72     ios::sync_with_stdio(false);
73     cin.tie(0);
74     cout.tie(0); 
75     solve();
76  
77     return 0;
78 }