Educational Codeforces Round 106 (Rated for Div. 2) B. Binary Removals 思维+字符串

• ​​原题链接​​​
• 测试样例

input
5
10101011011
0000
11111
110
1100
output
YES
YES
YES
YES
NO

• 
  
• Note

In the first testcase you can choose a sequence a=[1,3,6,9]. Removing the underlined letters from “10101011011” will produce a string “0011111”, which is sorted.
In the second and the third testcases the sequences are already sorted.
In the fourth testcase you can choose a sequence a=[3]. s′= “11”, which is sorted.
In the fifth testcase there is no way to choose a sequence a such that s′ is sorted.

• 
  
• 解题思路
  这道题其实就是一道理解题，我们知道，我们不能在连续的地方删除，所以倘若存在至少2个连续的1在至少2个连续的0左边，那么这个字符串怎么删都不能保证是非递减的。 所以我们只要判断是否存在这种情况即可。
• 代码

/**
* @filename:B.cbp
* @Author : pursuit
* @Blog:unique_pursuit
* @email: 2825841950@qq.com
* @Date : 2021-03-18-23.10.03
*/
#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
const int INF = 0x3f3f3f3f;
const int maxn=1e5+5;
const int mod=1e9+7;

//
int t;
string s;
void solve(){
    int len=s.size();
    int index0,index1;
    for(index1=0;index1<len-1;index1++){
        if(s[index1]=='1'&&s[index1+1]=='1'){
            break;
        }
    }
    for(index0=len-2;index0>=0;index0--){
        if(s[index0]=='0'&&s[index0+1]=='0'){
            break;
        }
    }
    if(index1<index0){
        cout<<"NO"<<endl;
    }
    else{
        cout<<"YES"<<endl;
    }
}
int main(){
    while(cin>>t){
        while(t--){
            cin>>s;
            solve();
        }
    }
    return 0;
}