给定字符串,字符串中的问号表示可以自定义字母,字符串中是否只有一个abacaba的子串
比赛时没有想到暴力,结束后发现这题可以暴力解决,直接从每一位开始枚举即可。不过这题暴力写着比较麻烦
#include<bits/stdc++.h>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
string pos="abacaba";
int Find(string str){
int cnt=0;
rep(i,0,str.size()-6){
int flag=1;
rep(j,0,7){
if(str[i+j]!=pos[j]){
flag=0;
break;
}
}
if(flag) cnt++;
}
return cnt;
}
void solve(){
int n;cin>>n;
string s;cin>>s;
int cnt=Find(s);
if(cnt==1){
cout<<"YES"<<endl;
rep(i,0,s.size()){
if(s[i]!='?') cout<<s[i];
else cout<<'z';
}
cout<<endl;
}else if(cnt>1){
cout<<"NO"<<endl;
}else{
string tmp=s;
int flag1=1;
rep(i,0,s.size()-6){
int flag=1;
s=tmp;
rep(j,0,7){
if(s[i+j]=='?') s[i+j]=pos[j];
if(s[i+j]!=pos[j]) flag=0;
}
if(flag){
cnt=Find(s);
if(cnt==1){
flag1=0;
cout<<"YES"<<endl;
rep(k,0,s.size()){
if(s[k]!='?') cout<<s[k];
else cout<<'z';
}
cout<<endl;
}
if(!flag1) break;
}
}
if(flag1) cout<<"NO"<<endl;
}
}
int main(){
IOS;
int t;cin>>t;
while(t--){
solve();
}
return 0;
}