https://codeforces.com/contest/1499/problem/B
Ideas:
The final answer is 00000, 11111, and 00000111 three states, and then these states are checked according to the structure.
The third state enumeration intermediate point i,[1,i]=0,[i+1,n]=1;
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=200;
typedef long long LL;
inline LL read(){LL x=0,f=1;char ch=getchar(); while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
char s[maxn];
int main(void)
{
cin.tie(0);std::ios::sync_with_stdio(false);
LL t;cin>>t;
while(t--){
cin>>(s+1);
LL len=strlen(s+1);
bool flag1=1;bool flag2=1;bool flag3=0;
///全0
LL pos=-1;
for(LL i=1;i<=len;i++){
if(s[i]=='1'&&i-pos>=2){
pos=i;
}
else if(s[i]=='1'&&i-pos<2){
flag1=0;
break;
}
}
///全1
pos=-1;
for(LL i=1;i<=len;i++){
if(s[i]=='0'&&i-pos>=2){
pos=i;
}
else if(s[i]=='0'&&i-pos<2){
flag2=0;
break;
}
}
///前0后1
for(LL i=1;i<=len;i++){///[1,i]为0,[i+1,len]为1
LL p=-1;
bool ok=1;
for(LL j=1;j<=i;j++){
if(s[j]=='1'&&j-p>=2){
p=j;
}
else if(s[j]=='1'&&j-p<2){
ok=0;
break;
}
}
if(ok==0) continue;
for(LL j=i+1;j<=len;j++){
if(s[j]=='0'&&j-p>=2){
p=j;
}
else if(s[j]=='0'&&j-p<2){
ok=0;
break;
}
}
if(ok==1){
flag3=1;
break;
}
}
if(flag1||flag2||flag3){
cout<<"YES"<<"\n";
}
else cout<<"NO"<<"\n";
}
return 0;
}