C - 4-adjacent


Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

We have a sequence of length N, a=(a1,a2,…,aN). Each ai is a positive integer.

Snuke's objective is to permute the element in a so that the following condition is satisfied:

  • For each 1≤iN−1, the product of ai and ai+1 is a multiple of 4.

Determine whether Snuke can achieve his objective.

Constraints

  • 2≤N≤105
  • ai is an integer.
  • 1≤ai≤109

Input

Input is given from Standard Input in the following format:

N
a1 a2 aN

Output

If Snuke can achieve his objective, print Yes; otherwise, print No.


Sample Input 1

3
1 10 100

Sample Output 1

Yes

One solution is (1,100,10).


Sample Input 2

4
1 2 3 4

Sample Output 2

No

It is impossible to permute a so that the condition is satisfied.


Sample Input 3

3
1 4 1

Sample Output 3

Yes

The condition is already satisfied initially.


Sample Input 4

2
1 1

Sample Output 4

No

Sample Input 5

6
2 7 1 8 2 8

Sample Output 5

Yes
1~n-1之间保证a[i]*a[i+1]%4==0
    #include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int a,b,c,x,n;
int main()
{
while(scanf("%d",&n)!=EOF)
{
a=b=c=;
for(int i=;i<n;i++)
{
scanf("%d",&x);
if(!(x%)) a++;
else if(x&) b++;
else c++;
}
if(!c) b--;
puts(a>=b?"Yes":"No");
}
return ;
}

D - Grid Coloring


Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

We have a grid with H rows and W columns of squares. Snuke is painting these squares in colors 1, 2, , N. Here, the following conditions should be satisfied:

  • For each i (1≤iN), there are exactly ai squares painted in Color i. Here, a1+a2+…+aN=HW.
  • For each i (1≤iN), the squares painted in Color i are 4-connected. That is, every square painted in Color i can be reached from every square painted in Color i by repeatedly traveling to a horizontally or vertically adjacent square painted in Color i.

Find a way to paint the squares so that the conditions are satisfied. It can be shown that a solution always exists.

Constraints

  • 1≤H,W≤100
  • 1≤NHW
  • ai≥1
  • a1+a2+…+aN=HW

Input

Input is given from Standard Input in the following format:

H W
N
a1 a2 aN

Output

Print one way to paint the squares that satisfies the conditions. Output in the following format:

c11  c1W
:
cH1 cHW

Here, cij is the color of the square at the i-th row from the top and j-th column from the left.


Sample Input 1

2 2
3
2 1 1

Sample Output 1

1 1
2 3

Below is an example of an invalid solution:

1 2
3 1

This is because the squares painted in Color 1 are not 4-connected.


Sample Input 2

3 5
5
1 2 3 4 5

Sample Output 2

1 4 4 4 3
2 5 4 5 3
2 5 5 5 3

Sample Input 3

1 1
1
1

Sample Output 3

1
h*w的网格,填充颜色,颜色种类为n,a[i]*****a[n],为每种颜色的个数,保证所填充相等颜色之间必须联通,蛇形填充就行。
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <cassert>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<=y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 1044266558
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
int g[][];
int h,w,n,x,k;
int main()
{
while(scanf("%d%d%d",&h,&w,&n)!=EOF)
{
mem(g);
k=-;
for(int i=;i<=n;i++)
{
scanf("%d",&x);
while(x--) k++,g[(k/h)&?(h--(k%h)):k%h][k/h]=i;
}
for(int i=;i<h;i++)
{
for(int j=;j<w;j++)
{
if(j) printf(" ");
printf("%d",g[i][j]);
}
printf("\n");
}
}
return ;
}

Atcoder ABC 069 C - 4-adjacent D - Grid Coloring的更多相关文章

  1. AtCoder ABC 042D いろはちゃんとマス目 / Iroha and a Grid

    题目链接:https://abc042.contest.atcoder.jp/tasks/arc058_b 题目大意: 给定一个 H * W 的矩阵,其中左下角 A * B 区域是禁区,要求在不踏入禁 ...

  2. AtCoder Regular Contest 080 D - Grid Coloring

    地址:http://arc080.contest.atcoder.jp/tasks/arc080_b 题目: D - Grid Coloring Time limit : 2sec / Memory ...

  3. ATCODER ABC 099

    ATCODER ABC 099 记录一下自己第一场AK的比赛吧...虽然还是被各种踩... 只能说ABC确实是比较容易. A 题目大意 给你一个数(1~1999),让你判断它是不是大于999. Sol ...

  4. Atcoder ABC 141

    Atcoder ABC 141 A - Weather Prediction SB题啊,不讲. #include<iostream> #include<cstdio> #inc ...

  5. Atcoder ABC 139E

    Atcoder ABC 139E 题意: n支球队大循环赛,每支队伍一天只能打一场,求最少几天能打完. 解法: 考虑抽象图论模型,既然一天只能打一场,那么就把每一支球队和它需要交手的球队连边. 求出拓 ...

  6. Atcoder ABC 139D

    Atcoder ABC 139D 解法: 等差数列求和公式,记得开 $ long long $ CODE: #include<iostream> #include<cstdio> ...

  7. Atcoder ABC 139C

    Atcoder ABC 139C 题意: 有 $ n $ 个正方形,选择一个起始位置,使得从这个位置向右的小于等于这个正方形的高度的数量最多. 解法: 简单递推. CODE: #include< ...

  8. Atcoder ABC 139B

    Atcoder ABC 139B 题意: 一开始有1个插口,你的插排有 $ a $ 个插口,你需要 $ b $ 个插口,问你最少需要多少个插排. 解法: 暴力模拟. CODE: #include< ...

  9. Atcoder ABC 139A

    Atcoder ABC 139A 题意: 给你两个字符串,记录对应位置字符相同的个数 $ (n=3) $ 解法: 暴力枚举. CODE: #include<iostream> #inclu ...

随机推荐

  1. mysql5.7 安装方法 (跟旧的不一样了)

    MySQL 5.7发布之后很多网友都在说,打开想安装文件夹.但是文件夹中没有DATA目录, 没有mysqly默认库.启动不了数据库,那是因为5.7的数据库的初始化方法和之前的初始化不一样了. 首先这里 ...

  2. "pom.xml" could not be activated because it does not exist.

    "pom.xml" could not be activated because it does not exist. 在sts中使用maven build,输入package然后 ...

  3. springMVC 配置jdbcTemplate连接Oracle数据库出错

    springMVC 配置jdbcTemplate连接Oracle数据库出错 错误信息: log4j:WARN No appenders could be found for logger (org.s ...

  4. 图像切割—基于图的图像切割(Graph-Based Image Segmentation)

     图像切割-基于图的图像切割(Graph-Based Image Segmentation) Reference: Efficient Graph-Based Image Segmentation ...

  5. 后缀自己主动机(SAM)学习指南

    *在学习后缀自己主动机之前须要熟练掌握WA自己主动机.RE自己主动机与TLE自己主动机* 什么是后缀自己主动机 后缀自己主动机 Suffix Automaton (SAM) 是一个用 O(n) 的复杂 ...

  6. [Perl系列—] 2. Perl 中的引用使用方法

    Perl 中的引用,为什么要使用引用? 对于熟悉C语言的开发人员来说, 指针这个概念一定不陌生. Perl 的引用就是指针,能够指向变量.数组.哈希表甚至子程序. Perl5中的两种Perl引用类型为 ...

  7. angularjs 指令2

    <!DOCTYPE HTML> <html ng-app="myApp"> <head> <meta http-equiv="C ...

  8. 手机表单验证插件mvalidate的使用

    使用 1.引入js和css <script type="text/javascript" src="../script/jquery-mvalidate.js&qu ...

  9. 6.C语言文件操作之英语电子字典的实现,dos版

    多的不说,直接上代码: 里面涉及的字典文件在这:这是传送门,下载下来以后把该文件放在工程目录下即可 #define _CRT_SECURE_NO_WARNINGS #include <stdio ...

  10. Android项目实战(五十五):部分机型点击home再点图标进入程序不保留再之前界面的问题

    解决办法: 1.在基类Activity中 添加方法 @Override public boolean moveTaskToBack(boolean nonRoot) { return super.mo ...