HDU - 6400 多校8 Parentheses Matrix(构造)
Parentheses Matrix
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
Special Judge
- 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.
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.
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAX = ;
typedef long long LL; int main(void)
{
int t,n,m,i,j;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
if((n&)&&(m&)){
for(i=;i<=n;i++){
for(j=;j<=m;j++){
printf("(");
}
printf("\n");
}
}
else if(n&){
for(i=;i<=n;i++){
for(j=;j<=m;j++){
if(j&) printf("(");
else printf(")");
}
printf("\n");
}
}
else if(m&){
for(i=;i<=n;i++){
for(j=;j<=m;j++){
if(i&) printf("(");
else printf(")");
}
printf("\n");
}
}
else{
if(n>=m){
for(i=;i<=n;i++){
for(j=;j<=m;j++){
if(i==&&(m/-)>){
printf("(");
continue;
}
if(i==n&&(m/-)>){
printf(")");
continue;
}
if(j==){
printf("(");
continue;
}
if(j==m){
printf(")");
continue;
}
if((i+j)&){
printf(")");
}
else{
printf("(");
}
}
printf("\n");
}
}
else{
for(i=;i<=n;i++){
for(j=;j<=m;j++){
if(j==&&(n/-)>){
printf("(");
continue;
}
if(j==m&&(n/-)>){
printf(")");
continue;
}
if(i==){
printf("(");
continue;
}
if(i==n){
printf(")");
continue;
}
if((i+j)&){
printf(")");
}
else{
printf("(");
}
}
printf("\n");
}
}
}
}
return ;
}
HDU - 6400 多校8 Parentheses Matrix(构造)的更多相关文章
- hdu 6400 Parentheses Matrix
题目链接 Problem Description A parentheses matrix is a matrix where every element is either '(' or ')'. ...
- hdu 2018多校8
A.Character Encoding 简单计数 m个非负数和等于k的方案数为$\binom{m+k-1}{k}$, 但题目还要求每个数小于n, 容斥一下即可 即$ans = \sum\limits ...
- hdu多校第八场Parentheses Matrix
#include<bits/stdc++.h> using namespace std; ][]; int main() { int t; scanf("%d",&am ...
- hdu 5015 233 Matrix(构造矩阵)
http://acm.hdu.edu.cn/showproblem.php?pid=5015 由于是个二维的递推式,当时没有想到能够这样构造矩阵.从列上看,当前这一列都是由前一列递推得到.依据这一点来 ...
- HDU 4671 Backup Plan (2013多校7 1006题 构造)
Backup Plan Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total ...
- HDU多校1003-Divide the Stones(构造)
Problem Description There are n stones numbered from 1 to n.The weight of the i-th stone is i kilogr ...
- HDU 3213 Box Relations(拓扑排序构造)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3231 题意:有n个长方体,四种限制条件.(1)I x y x和y有相交:(2)X/Y/Z x y x ...
- AGC027 D - Modulo Matrix 构造
目录 题目链接 题解 代码 题目链接 AGC027 D - Modulo Matrix 题解 从第左上角第一个点开始染色,相邻不同色,染法唯一 那么一个点的四周与他不同色,我们另这个点比四周都大,那么 ...
- Educational Codeforces Round 48 (Rated for Div. 2) D 1016D Vasya And The Matrix (构造)
D. Vasya And The Matrix time limit per test 2 seconds memory limit per test 256 megabytes input stan ...
随机推荐
- 九度OJ 1011:最大连续子序列 (DP)
时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:5615 解决:2668 题目描述: 给定K个整数的序列{ N1, N2, ..., NK },其任意连续子序列可表示为{ Ni, N ...
- Open-sourcing LogDevice, a distributed data store for sequential data
https://logdevice.io/blog/2018/09/12/open-sourcing-announcement.html September 12, 2018 We are exc ...
- 该 Bucket 已存在,或被其他用户占用
- JavaScript 四种显示数据方式
<!doctype html> <html lang="en"> <head> <meta charset="UTF-8&quo ...
- session,cookie的理解(总结)
会话(Session)跟踪是Web程序中常用的技术,用来跟踪用户的整个会话.常用的会话跟踪技术是Cookie与Session.Cookie通过在客户端记录信息确定用户身份,Session通过在服务器端 ...
- BZOJ 4519 [CQOI2016]不同的最小割
这道题目很奇怪. 为什么奇怪?因为这道题用了一种叫分治最小割/最小割树的玩意. 以前从来没有见过这东西. 推荐一个讲这玩意的博客 写起来还是很顺手的. #include<iostream> ...
- RQNOJ 622 最小重量机器设计问题:dp
题目链接:https://www.rqnoj.cn/problem/622 题意: 一个机器由n个部件组成,每一种部件都可以从m个不同的供应商处购得. w[i][j]是从供应商j处购得的部件i的重量, ...
- CI框架上传csv文件
今天遇到在用CI框架上传csv文件时报错问题: The filetype you are attempting to upload is not allowed. 是类型不允许,想到CI框架的conf ...
- codeforces 658B B. Bear and Displayed Friends(优先队列)
题目链接: B. Bear and Displayed Friends time limit per test 2 seconds memory limit per test 256 megabyte ...
- BZOJ3991:寻宝游戏 (LCA+dfs序+树链求并+set)
小B最近正在玩一个寻宝游戏,这个游戏的地图中有N个村庄和N-1条道路,并且任何两个村庄之间有且仅有一条路径可达.游戏开始时,玩家可以任意选择一个村庄,瞬间转移到这个村庄,然后可以任意在地图的道路上行走 ...