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()
#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;
}
};952. Largest Component Size by Common Factor | 212. Word Search II |
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123. Best Time to Buy and Sell Stock III | 85. Maximal Rectangle |
84. Largest Rectangle in Histogram | 60. Permutation Sequence |
42. Trapping Rain Water | 32. Longest Valid Parentheses |
Cutting a material | Bubble Sort |
Number of triangles | AND path in a binary tree |
Factorial equations | Removal of vertices |
Happy segments | Cyclic shifts |
Zoos | Build a graph |
Almost correct bracket sequence | Count of integers |
Differences of the permutations | Doctor's Secret |
Back to School | I am Easy |
Teddy and Tweety | Partitioning binary strings |
Special sets | Smallest chosen word |