Parentheses Matrix（构造）

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 #define ll long longusing 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;}

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