Parentheses Matrix（构造）

Problem Description

A parentheses matrix is a matrix where every element is either '(' or ')'. We define the goodness of a parentheses matrix as the number of balanced rows (from left to right) and columns (from up to down). Note that:

- an empty sequence is balanced;
- if A is balanced, then (A) is also balanced;
- if A and B are balanced, then AB is also balanced.

For example, the following parentheses matrix is a 2×4 matrix with goodness 3, because the second row, the second column and the fourth column are balanced:

)()(
()()

Now, give you the width and the height of the matrix, please construct a parentheses matrix with maximum goodness.

Input

The first line of input is a single integer T (1≤T≤50), the number of test cases.

Each test case is a single line of two integers h,w (1≤h,w≤200), the height and the width of the matrix, respectively.

Output

For each test case, display h lines, denoting the parentheses matrix you construct. Each line should contain exactly w characters, and each character should be either '(' or ')'. If multiple solutions exist, you may print any of them.

Sample Input

3 1 1 2 2 2 3

Sample Output

( () )( ((( )))

 

构造题，遇事不决就打表。

需要打表找规律，才能找出题解的规律。

代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e5+10;
const double INF=0x3f3f3f3f;
char ma[220][220];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        memset(ma,' ',sizeof(ma));
        scanf("%d%d",&n,&m);
        if((n&1)&&(m&1))
        {
            for(int i=1; i<=n; i++)
            {
                for(int j=1; j<=m; j++)
                {
                    printf("(");
                }
                printf("\n");
            }
        }
        if(!(n&1)&&(m&1))
        {
            for(int i=1; i<=n; i++)
            {
                for(int j=1; j<=m; j++)
                {
                    if(i&1)printf("(");
                    else printf(")");
                }
                printf("\n");
            }
        }
        if((n&1)&&!(m&1)) 
        {
            for(int i=1; i<=n; i++)
            {
                for(int j=1; j<=m; j+=2)
                {
                    printf("()");
                }
                printf("\n");
            }
        }
        if(!(n&1)&&!(m&1))//都是偶数
        {
            if(n<6||m<6) //特殊讨论
            {
                if(n<=m)
                {
                    for(int i=1; i<=n; i++)
                    {
                        for(int j=1; j<=m; j++)
                        {
                            if(i==1)ma[i][j]='(';
                            else if(i==n)ma[i][j]=')';
                            else if((i&1)==0)
                            {
                                if(j<=(m/2))ma[i][j]=')';
                                else ma[i][j]='(';
                            }
                            else
                            {
                                if(j<=(m/2))ma[i][j]='(';
                                else ma[i][j]=')';
                            }
                        }
                    }

                }
                else
                {
                    for(int i=1;i<=m;i++)
                    {
                        for(int j=1;j<=n;j++)
                        {
                            if(i==1)ma[j][i]='(';
                            else if(i==m)ma[j][i]=')';
                            else if((i&1)==0)
                            {
                                if(j<=(n/2))ma[j][i]=')';
                                else ma[j][i]='(';
                            }
                            else
                            {
                                if(j<=(n/2))ma[j][i]='(';
                                else ma[j][i]=')';
                            }
                        }
                    }
                }
                for(int i=1; i<=n; i++)
                {
                    for(int j=1; j<=m; j++)
                    {
                        printf("%c",ma[i][j]);
                    }
                    printf("\n");
                }
                continue;
            }
            for(int i=1; i<=m; i++)ma[1][i]='(';
            for(int i=1; i<=n; i++)ma[i][1]='(';
            for(int i=1; i<=m; i++)ma[n][i]=')';
            for(int i=2; i<=n; i++)ma[i][m]=')';
            for(int i=2; i<=n-1; i++)
            {
                for(int j=2; j<=m-1; j+=2)
                {
                    if((i&1)==0)
                    {
                        ma[i][j]=')';
                        ma[i][j+1]='(';
                    }
                    else
                    {
                        ma[i][j]='(';
                        ma[i][j+1]=')';
                    }
                }
            }
            for(int i=1; i<=n; i++)
            {
                for(int j=1; j<=m; j++)
                {
                    printf("%c",ma[i][j]);
                }
                printf("\n");
            }
        }
    }
    return 0;
}