Educational Codeforces Round 65 (Rated for Div. 2)(ACD)B是交互题，不怎么会

A. Telephone Number

A telephone number is a sequence of exactly 11 digits, where the first digit is 8. For example, the sequence 80011223388 is a telephone number, but the sequences 70011223388 and 80000011223388 are not.

You are given a string

In one operation you can delete any character from string

You need to determine whether there is such a sequence of operations (possibly empty), after which the string

Input

The first line contains one integer

The first line of each test case contains one integer

The second line of each test case contains the string

Output

For each test print one line.

If there is a sequence of operations, after which

Otherwise, print NO.

Example

input

Copy

2
13
7818005553535
11
31415926535

output

Copy

YES
NO

Note

In the first test case you need to delete the first and the third digits. Then the string 7818005553535 becomes 88005553535.

这个分一下是否满足11位，如果大于11位，只需要判断除去后10位前面是否有8即可

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
#include<vector>

const int maxn=1e5+5;
typedef long long ll;
using namespace std;

int main()
{
    int T;
    cin>>T;
    string str;
    int n;
    while(T--)
    {
        cin>>n;
    cin>>str;
    int len=str.length();
    if(len>11)
    {
        int flag=0;
        for(int t=len-11;t>=0;t--)
        {
            if(str[t]=='8')
            {
            flag=1;
            break;    
            }
        }
        if(flag)
        {
            cout<<"YES"<<endl;
        }
        else
        {
            cout<<"NO"<<endl;
        }
    }    
    else if(len==11)
    {
        if(str[0]=='8')
        {
            cout<<"YES"<<endl;
        }
        else
        {
            cout<<"NO"<<endl;
        }
    }
    else
    {
        cout<<"NO"<<endl;
    }
    }
    return 0;
}

C. News Distribution

In some social network, there are

Initially, some user

For each user

Input

The first line contains two integers

Then

It is guaranteed that

Output

Example

input

Copy

output

Copy

4 4 1 4 4 2 2 
裸的并查集
代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>

const int maxn=5e5+5;
typedef long long ll;
using namespace std;
int pre[maxn];
int find(int x)
{
    if(x==pre[x])
    {
        return x;
    }
    else
    {
        return pre[x]=find(pre[x]);
    }
}
void Merge(int x,int y)
{
    int xx=find(x);
    int yy=find(y);
    if(xx!=yy)
    {
        pre[xx]=yy;
    }
    
}
int a[maxn];
int main()
{
    int n,m;
    cin>>n>>m;
    int x;
    for(int t=1;t<=n;t++)
    {
        pre[t]=t;
    }
    int s1,s2;
    for(int t=0;t<m;t++)
    {
        scanf("%d",&x);
        if(x!=0)
        scanf("%d",&s1);
        for(int j=1;j<x;j++)
        {
            scanf("%d",&s2);
            Merge(s1,s2);
            s1=s2;
        }
    }

    int ans=0;
    memset(a,0,sizeof(a));
    for(int t=1;t<=n;t++)
    {
    
        a[find(t)]++;
    }
    for(int t=1;t<=n;t++)
    {
        printf("%d ",a[find(t)]);
    }

    
    return 0;
}

A string is called bracket sequence if it does not contain any characters other than "(" and ")". A bracket sequence is called regular(shortly, RBS) if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, "", "(())" and "()()" are RBS and ")(" and "(()" are not.

We can see that each opening bracket in RBS is paired with some closing bracket, and, using this fact, we can define nesting depth of the RBS as maximum number of bracket pairs, such that the

Now, you are given RBS

Among all possible variants you should choose one that minimizes maximum of

Input

The first line contains an even integer

The second line contains regular bracket sequence

Output

Print single string

Examples

input

Copy

2
()

output

Copy

input

Copy

4
(())

output

Copy

input

Copy

10
((()())())

output

Copy

0110001111

Note

In the first example one of optimal solutions is

In the second example it's optimal to make

In the third example we can make

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=2e5+5;
typedef long long ll;
using namespace std;
char str[maxn];
int main()
{
   int n;
   cin>>n;
   scanf("%s",str);
   int s1=0,s2=0;
   int a[maxn];
   for(int t=0;t<n;t++)
   {
       if(str[t]=='(')
       {
           if(s1<s2)
           {
               s1++;
               a[t]=0;
        }
        else
        {
            s2++;
            a[t]=1;
        }
    }
    else
    {
        if(s1<s2)
        {
            s2--;
            a[t]=1;
        }
        else
        {
            s1--;
            a[t]=0;
        }
    }
   }
   for(int t=0;t<n;t++)
   {
       printf("%d",a[t]);
   }
   return 0;
}

Educational Codeforces Round 65 (Rated for Div. 2)(ACD)B是交互题，不怎么会

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