1379A - Acacius and String - CodeForces Solution

brute force implementation strings *1500

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Python Code: 

T = "abacaba"

def count(s):
    i = 0
    j = 7
    c = 0
    while i < len(s) and j >= 0:
        if s[i:j] == "abacaba":
            c += 1
            i+=1
            j+=1
        else:
            i+=1
            j+=1
    return c

def solve():
    n = int(input())
    s = input()
    F = False

    if count(s) == 1:
        F = True
        print("Yes")
        sc = s.replace('?', 'd')
        print(sc)
        return

    for i in range(n - len(T) + 1):
        sc = list(s)
        ok = True
        for j in range(len(T)):
            if sc[i + j] != '?' and sc[i + j] != T[j]:
                ok = False
                break
            sc[i + j] = T[j]

        if ok and count(''.join(sc)) == 1:
            sc = ''.join(sc)
            sc = sc.replace('?', 'd')
            F = True
            print("Yes")
            print(sc)
            return

    if not F:
        print("No")
        return

t = int(input())
for _ in range(t):
    solve()

C++ Code: 

#include <bits/stdc++.h>
using namespace std;
#define ios ios::sync_with_stdio(0)
#define endl '\n'
#define int long long
#define ar array<int, 2>
#define arr array<int, 3>
const int N = 2e5 + 5, M = 2 * N;
const int inf = 0x3f3f3f3f;
int mod = 998244353; //1e9+7;
int t, n, m, k;
string a = "abacaba";
signed main()
{
    ios;
#ifdef DEBUG
    freopen("../1.in", "r", stdin);
#endif
    //  核心技巧在于那个tmp的 再验证。。的手法。。
    // 就是匹配这个string  abacaba
    // 问 是否能实现当且仅当 匹配一次。。
    // 刚开始还想到dfs
    // 实际上只要从左往右 一截一截的去匹配验证就可以了
    // 1 匹配。2 验证是否能够唯一。
    cin >> t;
    while (t--)
    {
        cin >> n;
        string s;
        cin >> s;
        string ans;
        for (int i = 0; i <= n - 7; ++i)
        {
            string tmp = s;
            bool f = 1;
            for (int j = 0; j < 7; ++j)
            {
                if (tmp[i + j] != '?' && tmp[i + j] != a[j])
                {
                    f = 0;
                    break;
                }
                else
                    tmp[i + j] = a[j];
            }
            if (!f)
                continue;
            for (int i = 0; i <n; ++i)
                if (tmp[i] == '?')
                    tmp[i] = 'z';
            for (int j = 0; j <= n - 7; ++j)
                if (i != j && tmp.substr(j, 7) == a)
                {
                    f = 0;
                    break;
                }
            if (f)
            {
                ans = tmp;
                break;
            }
        }

        cout << (ans.size() ? "YES" : "NO") << endl;
        if (ans.size())
            cout << ans << endl;
    }
};


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