CF 1379 A. Acacius and String

传送门

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。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
#include <cmath>
 
using namespace std;
 
#define ll long long
#define pb push_back
#define fi first
#define se second

const int N = 2e5 + 10;
char s[N];
int len;

void solve()
{      
    string p = "abacaba";
    int T;
    cin >> T;
    while(T--){
        string s, tmp;
        int n;
        cin >> n >> s;

        int good = 0;
        for(int i = 0; i < n; ++i){
            if(i + 6 >= n) break;

            //几个字符匹配
            int ok = 0;
            for(int j = 0; j < 7; ++j){
                if(s[i + j] == '?' || p[j] == s[i + j]) ok++;
                else break;
            }
            if(ok == 7){
                tmp = s;
                for(int j = 0; j < 7; ++j) tmp[i + j] = p[j];

                //出现几次    
                int cnt = 0;
                string::size_type inx = 0;
                while(1){
                    inx = tmp.find(p, inx);
                    //printf("inx = %d
", (int)inx);
                    if(inx == tmp.npos) break;
                    inx = inx + 3;
                    cnt++;
                }
                if(cnt == 1){
                    good = 1;
                    cout << "yes" << endl;
                    for(int j = 0; j < n; ++j){
                        cout << (tmp[j] == '?' ? 'z' : tmp[j]);
                    }
                    cout << endl;
                    break;
                }
            }
        }

        if(!good) cout << "no" << endl;
    }
}
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0); 
    solve();
 
    return 0;
}