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 $\mathrm{nn bracket\; sequences s1,s2,\dots ,sns1,s2,\dots ,sn.\; Calculate\; the\; number\; of\; pairs i,j(1\le i,j\le n)i,j(1\le i,j\le n) such\; that\; the\; bracket\; sequence si+sjsi+sj is\; a\; regular\; bracket\; sequence.\; Operation ++ means\; concatenation\; i.e.\; "()("\; +\; ")()"\; =\; "()()()".}$

If ${\mathrm{si+sjsi+sj and sj+sisj+si are\; regular\; bracket\; sequences\; and i\ne ji\ne j,\; then\; both\; pairs (i,j)(i,j) and (j,i)(j,i) must\; be\; counted\; in\; the\; answer.\; Also,\; if si+sisi+si is\; a\; regular\; bracket\; sequence,\; the\; pair (i,i)(i,i) must\; be\; counted\; in\; the\; answer.}}^{}$

Input

The first line contains one integer $\mathrm{n(1\le n\le 3\cdot 105)n(1\le n\le 3\cdot 105) —\; the\; number\; of\; bracket\; sequences.\; The\; following nn lines\; contain\; bracket\; sequences\; — non-empty strings\; consisting\; only\; of\; characters\; "("\; and\; ")".\; The\; sum\; of\; lengths\; of\; all\; bracket\; sequences\; does\; not\; exceed 3\cdot 1053\cdot 105.}$

Output

In the single line print a single integer — the number of pairs $\mathrm{i,j(1\le i,j\le n)i,j(1\le i,j\le n) such\; that\; the\; bracket\; sequence si+sjsi+sj is\; a\; regular\; bracket\; sequence.}$

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+50,mod = 998244353,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=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%s",c);
        int len=strlen(c);
        a[i].l=a[i].r=0;
        for(int j=0;j<len;j++)
        {
            //cout<<i<<" "<<c[i]<<endl;
            if(c[j]=='(')
                a[i].l++;
            else if(c[j]==')'&&a[i].l==0)
                a[i].r++;
            else
                a[i].l--;
        }
    }
    for(int i=0;i<n;i++)
    {
        if(a[i].l==0&&a[i].r==0)
            m++;
        else if(a[i].l==0)
            mr[a[i].r]++;
        else if(a[i].r==0)
            ml[a[i].l]++;
    }
    ll ans=0;
    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-1)+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 $\mathrm{n,a,bn,a,b.\; You\; need\; to\; find\; an\; adjacency\; matrix\; of\; such\; an\; undirected\; graph\; that\; the\; number\; of\; components\; in\; it\; is\; equal\; to aa,\; and\; the\; number\; of\; components\; in\; its\; complement\; is bb.\; The\; matrix\; must\; be\; symmetric,\; and\; all\; digits\; on\; the\; main\; diagonal\; must\; be\; zeroes.}$

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

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

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

Input

In a single line, three numbers are given $\mathrm{n,a,b(1\le n\le 1000,1\le a,b\le n)n,a,b(1\le n\le 1000,1\le a,b\le n):\; is\; the\; number\; of\; vertexes\; of\; the\; graph,\; the\; required\; number\; of\; connectivity\; components\; in\; it,\; and\; the\; required\; amount\; of\; the\; connectivity\; component\; in\; it\text{'}s\; complement.}$

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 $\mathrm{nn lines,\; output nn digits\; such\; that jj-th\; digit\; of ii-th\; line\; must\; be 11if\; and\; only\; if\; there\; is\; an\; edge\; between\; vertices ii and jj in GG (and 00 otherwise).\; Note\; that\; the\; matrix\; must\; be\; symmetric,\; and\; all\; digits\; on\; the\; main\; diagonal\; must\; be\; zeroes.}$

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+50,mod = 998244353,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!=1&&b!=1)
    {
        cout<<"NO"<<endl;
    }
    else if(a==1&&b==1)
    {
        if(n==3||n==2)
            cout<<"NO"<<endl;
        else
        {
            for(int i=1;i<=n;i++)
            {
                g[i][i+1]=1;
                g[i+1][i]=1;
            }
            cout<<"YES"<<endl;
            for(int i=1;i<=n;i++)
            {
                for(int j=1;j<=n;j++)
                    printf("%d",g[i][j]);
                printf("
");
            }
        }
    }
    else
    {
        int temp=max(a,b)-1;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(i==j)
                    g[i][j]=0;
                else if(i<=temp||j<=temp)
                    g[i][j]=0;
                else
                    g[i][j]=1;
            }
        }
        if(a==1)
        {
            for(int i=1;i<=n;i++)
                for(int j=1;j<=n;j++)
                    if(i==j)
                        g[i][j]=0;
                    else
                        g[i][j]=!g[i][j];
        }
        printf("YES
");
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
                printf("%d",g[i][j]);
            printf("
");
        }
    }
}