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. 紫书 例题 11-2 UVa 1395(最大边减最小边最小的生成树)

    思路:枚举所有可能的情况. 枚举最小边, 然后不断加边, 直到联通后, 这个时候有一个生成树.这个时候,在目前这个最小边的情况可以不往后枚举了, 可以直接更新答案后break. 因为题目求最大边减最小 ...

  2. STM32为什么必须先配置时钟

    首先,任何外设都需要时钟,51单片机,stm32,430等等,因为寄存器是由D触发器组成的,往触发器里面写东西,前提条件是有时钟输入. 51单片机不需要配置时钟,是因为一个时钟开了之后所有的功能都可以 ...

  3. 国庆 day 7 下午

    思路:见博客. #include<iostream> #include<cstdio> #include<cstring> #include<algorith ...

  4. cocos2dx下的A星算法

    这是我依据这篇博文http://hi.baidu.com/wsapyoemdfacmqr/item/bdfb5c0a74c904d01ef0466d.来在cocos2dx上编写.这是终于的效果图: 红 ...

  5. rest_framework 分页三种

    .分页 a. 分页 看第n页 每页显示n条数据: b. 分页 在某个位置 向后查看多少条数据 c. 加密分页 上一页和下一页 本质:查看 记住页码id的最大值和最小值 通过其来准确扫描 过去的话 会从 ...

  6. centos7 安装swftools Apache_OpenOffice

    centos7 yum -y install wget wget http://www.swftools.org/swftools-0.9.2.tar.gz tar -xf swftools-.tar ...

  7. jqueryEasyUI form表单提交的一个困惑

    今天用到了jqueryEasyUI的form表单做一个增加操作的提交,想打开调试(用的是火狐)看看传的参数,但是怎么也看不到form表单提交的http请求?而且还会发送一个另外的请求! 在页面加载时, ...

  8. ipad mini2 升级9.0.2后解锁白屏解决

    解锁白屏是个什么现象?就是当你用手指滑动解锁后出现输入密码的界面后,1秒之内屏幕变白,中间一个黑色的苹果,几秒之后重新回到滑动解锁的界面.我出现这个现象不是因为升级了9.0.2,而是升级了9.0.2之 ...

  9. [POI2015]WIL-Wilcze doły(单调队列)

    题意 给定一个长度为n的序列,你有一次机会选中一段连续的长度不超过d的区间,将里面所有数字全部修改为0.请找到最长的一段连续区间,使得该区间内所有数字之和不超过p. (1<=d<=n< ...

  10. CF85E Guard Towers(二分答案+二分图)

    题意 已知 N 座塔的坐标,N≤5000 把它们分成两组,使得同组内的两座塔的曼哈顿距离最大值最小 在此前提下求出有多少种分组方案 mod 109+7 题解 二分答案 mid 曼哈顿距离 >mi ...