Educational Codeforces Round 45 (Rated for Div. 2) C、D

 

C. Bracket Sequences Concatenation Problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A bracket sequence is a string containing only characters "(" and ")".

A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

You are given nn bracket sequences s1,s2,…,sns1,s2,…,sn. Calculate the number of pairs i,j(1≤i,j≤n)i,j(1≤i,j≤n) such that the bracket sequence si+sjsi+sj is a regular bracket sequence. Operation ++ means concatenation i.e. "()(" + ")()" = "()()()".

If si+sjsi+sj and sj+sisj+si are regular bracket sequences and i≠ji≠j, then both pairs (i,j)(i,j) and (j,i)(j,i) must be counted in the answer. Also, if si+sisi+si is a regular bracket sequence, the pair (i,i)(i,i) must be counted in the answer.

Input

The first line contains one integer n(1≤n≤3⋅105)n(1≤n≤3⋅105) — the number of bracket sequences. The following nn lines contain bracket sequences — non-empty strings consisting only of characters "(" and ")". The sum of lengths of all bracket sequences does not exceed 3⋅1053⋅105.

Output

In the single line print a single integer — the number of pairs i,j(1≤i,j≤n)i,j(1≤i,j≤n) such that the bracket sequence si+sjsi+sj is a regular bracket sequence.

Examples

input

Copy

3
)
()
(

output

Copy

2

input

Copy

2
()
()

output

Copy

4

Note

In the first example, suitable pairs are (3,1)(3,1) and (2,2)(2,2).

In the second example, any pair is suitable, namely (1,1),(1,2),(2,1),(2,2)(1,1),(1,2),(2,1),(2,2).

http://codeforces.com/contest/990/problem/C

题意 n个括号字符串 有多少种组合把两个字符串连起来之后还是合法字符串   若（i,j）(j,i) 都合法 且i != j 算两种不同组合 反之算一种

解析  对每个字符串的未能在本字符串匹配的左右括号进行统计 比如(())  左右都是0 (() 右是1  左使0 )((( 右是3 左是1   再根据左右括号数量匹配

对于一个字符串只有左右括号至少有一个为零的才能进行匹配

1 左括号为零的和右括号为零进行匹配

2 左右括号都为零的进行匹配

AC代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e5+,mod = ,inf=0x3f3f3f3f;
typedef long long ll;
struct node
{
    int l,r;
}a[maxn];
char c[maxn];
int main()
{
    map<int,int> ml,mr;
    int n;
    ll m=;
    scanf("%d",&n);
    for(int i=;i<n;i++)
    {
        scanf("%s",c);
        int len=strlen(c);
        a[i].l=a[i].r=;
        for(int j=;j<len;j++)
        {
            //cout<<i<<" "<<c[i]<<endl;
            if(c[j]=='(')
                a[i].l++;
            else if(c[j]==')'&&a[i].l==)
                a[i].r++;
            else
                a[i].l--;
        }
    }
    for(int i=;i<n;i++)
    {
        if(a[i].l==&&a[i].r==)
            m++;
        else if(a[i].l==)
            mr[a[i].r]++;
        else if(a[i].r==)
            ml[a[i].l]++;
    }
    ll ans=;
    for(auto it=ml.begin();it!=ml.end();it++)
    {
        int u=it->first;
       // cout<<ans<<" "<<u<<endl;
        if(mr[u])
            ans+=(ll)ml[u]*mr[u];
    }
    cout<<ans+m*(m-)+m<<endl;
}

D. Graph And Its Complement

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Given three numbers n,a,bn,a,b. You need to find an adjacency matrix of such an undirected graph that the number of components in it is equal to aa, and the number of components in its complement is bb. The matrix must be symmetric, and all digits on the main diagonal must be zeroes.

In an undirected graph loops (edges from a vertex to itself) are not allowed. It can be at most one edge between a pair of vertices.

The adjacency matrix of an undirected graph is a square matrix of size nn consisting only of "0" and "1", where nn is the number of vertices of the graph and the ii-th row and the ii-th column correspond to the ii-th vertex of the graph. The cell (i,j)(i,j) of the adjacency matrix contains 11if and only if the ii-th and jj-th vertices in the graph are connected by an edge.

A connected component is a set of vertices XX such that for every two vertices from this set there exists at least one path in the graph connecting this pair of vertices, but adding any other vertex to XX violates this rule.

The complement or inverse of a graph GG is a graph HH on the same vertices such that two distinct vertices of HH are adjacent if and only if they are not adjacent in GG.

Input

In a single line, three numbers are given n,a,b(1≤n≤1000,1≤a,b≤n)n,a,b(1≤n≤1000,1≤a,b≤n): is the number of vertexes of the graph, the required number of connectivity components in it, and the required amount of the connectivity component in it's complement.

Output

If there is no graph that satisfies these constraints on a single line, print "NO" (without quotes).

Otherwise, on the first line, print "YES"(without quotes). In each of the next nn lines, output nn digits such that jj-th digit of ii-th line must be 11if and only if there is an edge between vertices ii and jj in GG (and 00 otherwise). Note that the matrix must be symmetric, and all digits on the main diagonal must be zeroes.

If there are several matrices that satisfy the conditions — output any of them.

Examples

input

Copy

3 1 2

output

Copy

YES
001
001
110

input

Copy

3 3 3

output

Copy

NO
http://codeforces.com/contest/990/problem/D题意 n个点的无向图 为是否存在某种 原图的连通分量为a 其补图的连通分量为b (且该原图的邻接矩阵是对称的 没有自环 其实只是个没用的条件 因为 无向图的邻接矩肯定是对称的）
解析 思考一下 就可以得出 a和b 至少要有一个为1 且a==1&&b==1时 要特判 尤其是1 被坑了一手

AC代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e3+,mod = ,inf=0x3f3f3f3f;
typedef long long ll;
int g[maxn][maxn];
int main()
{
    int n,a,b;
    scanf("%d %d %d",&n,&a,&b);
    if(a!=&&b!=)
    {
        cout<<"NO"<<endl;
    }
    else if(a==&&b==)
    {
        if(n==||n==)
            cout<<"NO"<<endl;
        else
        {
            for(int i=;i<=n;i++)
            {
                g[i][i+]=;
                g[i+][i]=;
            }
            cout<<"YES"<<endl;
            for(int i=;i<=n;i++)
            {
                for(int j=;j<=n;j++)
                    printf("%d",g[i][j]);
                printf("\n");
            }
        }
    }
    else
    {
        int temp=max(a,b)-;
        for(int i=;i<=n;i++)
        {
            for(int j=;j<=n;j++)
            {
                if(i==j)
                    g[i][j]=;
                else if(i<=temp||j<=temp)
                    g[i][j]=;
                else
                    g[i][j]=;
            }
        }
        if(a==)
        {
            for(int i=;i<=n;i++)
                for(int j=;j<=n;j++)
                    if(i==j)
                        g[i][j]=;
                    else
                        g[i][j]=!g[i][j];
        }
        printf("YES\n");
        for(int i=;i<=n;i++)
        {
            for(int j=;j<=n;j++)
                printf("%d",g[i][j]);
            printf("\n");
        }
    }
}