Atcoder ABC 069 C - 4-adjacent D - Grid Coloring

Posted 2020-10-05 十年换你一句好久不见

C - 4-adjacent

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Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

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

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

• For each 1≤i≤N?1, the product of a_i and a_i+1 is a multiple of 4.

Determine whether Snuke can achieve his objective.

Constraints

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

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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.

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Sample Input 1

3
1 10 100

Sample Output 1

Yes

One solution is (1,100,10).

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Sample Input 2

4
1 2 3 4

Sample Output 2

No

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

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Sample Input 3

3
1 4 1

Sample Output 3

Yes

The condition is already satisfied initially.

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Sample Input 4

2
1 1

Sample Output 4

No

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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=0;
            for(int i=0;i<n;i++)
            {
                scanf("%d",&x);
                if(!(x%4)) a++;
                else if(x&1) b++;
                else c++;
            }
            if(!c) b--;
            puts(a>=b?"Yes":"No");
        }
        return 0;
    }

 

D - Grid Coloring

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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≤i≤N), there are exactly a_i squares painted in Color i. Here, a₁+a₂+…+a_N=HW.
• For each i (1≤i≤N), 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≤N≤HW
• a_i≥1
• a₁+a₂+…+a_N=HW

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

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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.

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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

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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[110][110];
int h,w,n,x,k;
int main()
{
    while(scanf("%d%d%d",&h,&w,&n)!=EOF)
    {
        mem(g);
        k=-1;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            while(x--) k++,g[(k/h)&1?(h-1-(k%h)):k%h][k/h]=i;
        }
        for(int i=0;i<h;i++)
        {
            for(int j=0;j<w;j++)
            {
                if(j) printf(" ");
                printf("%d",g[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}