HDU 多校 6400 Parentheses Matrix(构造)

Posted 2021-12-24 Harris-H

HDU 多校 6400 Parentheses Matrix(构造)

// Problem: D. Parentheses Matrix
// Contest: Codeforces - 2018 Chinese Multi-University Training, Nanjing U Contest
// URL: https://codeforces.ml/gym/102192/problem/D
// Memory Limit: 128 MB
// Time Limit: 1000 ms
// Date: 2021-11-29 19:27:17
// --------by Herio--------

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull; 
const int N=1e3+5,M=2e4+5,inf=0x3f3f3f3f,mod=1e9+7;
const int hashmod[4] = 402653189,805306457,1610612741,998244353;
#define mst(a,b) memset(a,b,sizeof a)
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define x first
#define y second
#define pb emplace_back
#define SZ(a) (int)a.size()
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
#define ios ios::sync_with_stdio(false),cin.tie(nullptr) 
void Print(int *a,int n)
	for(int i=1;i<n;i++)
		printf("%d ",a[i]);
	printf("%d\\n",a[n]); 

template <typename T>		//x=max(x,y)  x=min(x,y)
void cmx(T &x,T y)
	if(x<y) x=y;

template <typename T>
void cmn(T &x,T y)
	if(x>y) x=y;

char a[N][N];
int main()
	int t;cin>>t;
	while(t--)
		int n,m;cin>>n>>m;
		if( (n%2==1) && (m%2==1) )
			rep(i,1,n)
				rep(j,1,m) putchar('(');
				putchar('\\n');
			
			continue;
		
		else if( (m%2==1)  )
			rep(i,1,n)
				if(i&1) rep(j,1,m) putchar('(');
				else rep(j,1,m) putchar(')');
				putchar('\\n');
			
		
		else if( (n%2==1) ) 
			rep(i,1,n)
				rep(j,1,m) putchar(j&1?'(':')');
				putchar('\\n');
						
		
		else 
			bool rev = 0;
			if(n > m) swap(n,m),rev = true;
			if( n == 2)
				rep(i,1,m) a[1][i]='(',a[2][i]=')'; 
			
			else if(n==4)
				rep(i,1,n)
					
						rep(j,1,m/2) a[i][j]=i&1?'(':')';
						rep(j,m/2+1,m) a[i][j]=i<=n/2?'(':')';
					
			
			else 
				rep(i,1,n)
					rep(j,1,m)
						if(i==1||j==1) a[i][j]='(';
						else if(i==n||j==m) a[i][j]=')';
						else if((i+j)&1) a[i][j]='(';
						else a[i][j]=')';
					
				
			
			if(rev)
				//printf("n=%d m=%d\\n",n,m);
				rep(i,1,m)
					rep(j,1,n) putchar(a[j][i]);
					putchar('\\n');
				
			
			else 
				//printf("n=%d m=%d\\n",n,m);
				rep(i,1,n)
					rep(j,1,m) putchar(a[i][j]);
					putchar('\\n');
				
			
		
	
	return 0;