AtCoder Regular Contest 080

手贱去开了abc，这么无聊。直接arc啊

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≤i≤N−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

Copy

3

1 10 100

Sample Output 1

Copy

Yes

One solution is (1,100,10).

Sample Input 2

Copy

4

1 2 3 4

Sample Output 2

Copy

No

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

Sample Input 3

Copy

3

1 4 1

Sample Output 3

Copy

Yes

The condition is already satisfied initially.

Sample Input 4

Copy

2

1 1

Sample Output 4

Copy

No

Sample Input 5

Copy

6

2 7 1 8 2 8

Sample Output 5

Copy

Yes

给你一个序列，通过你的安排让这个序列满足相邻两个数的乘积是4的倍数，如果有奇数的话,而且4的倍数个数+1还比奇数多，没有2的倍数，比如1 4 1就是成立的，

2的倍数很多，这些放哪都行，只要奇数和4的倍数能凑就行

#include <bits/stdc++.h>

using namespace std;

const int N=1e5+;

int a[N];

int main()

{

    int n;

    cin>>n;

    int cnt4=,cnt2=;

    for(int i=; i<n; i++)

    {

        cin>>a[i];

        a[i]%=;

        if(a[i]==)

            cnt4++;

        else if(a[i]==)

            cnt2++;

    }

    int cnt=n-cnt2-cnt4;

    if(cnt2)

    {

        if(cnt4>=cnt){

            cout<<"Yes";

            return ;

        }

    }

    else

    {

        if(cnt4&&(cnt4+>=cnt)){

            cout<<"Yes";

            return ;

        }

    }

    cout<<"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≤i≤N), there are exactly ai squares painted in Color i. Here, a1+a2+…+aN=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
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

Copy

Sample Output 1

Copy

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

Copy

Sample Output 2

Copy

Sample Input 3

Copy

Sample Output 3

Copy

我的做法莫名wa，换了个做法ac，也差不到原因，反正注意下就好

就是给你n个数，代表第几个数有几个，他们必须是联通的，特判题

#include <bits/stdc++.h>

using namespace std;

vector<int>a[];

int main()

{

    int h,w;

    cin>>h>>w;

    int n;

    cin>>n;

    int f=;

    for(int i=; i<n; i++)

    {

        int x;

        scanf("%d",&x);

        for(int j=;j<x;j++)

            a[f/w].push_back(i+),f++;

    }

    for(int i=; i<h; i++)

    {

        if(i&)reverse(a[i].begin(),a[i].end());

        printf("%d",a[i][]);

        for(int j=; j<w; j++)

            printf(" %d",a[i][j]);

        printf("\n");

    }

    return ;

}

E - Young Maids

Time limit : 2sec / Memory limit : 256MB

Score : 800 points

Problem Statement

Let N be a positive even number.

We have a permutation of (1,2,…,N), p=(p1,p2,…,pN). Snuke is constructing another permutation of (1,2,…,N), q, following the procedure below.

First, let q be an empty sequence. Then, perform the following operation until p becomes empty:

Select two adjacent elements in p, and call them x and y in order. Remove x and y from p (reducing the length of p by 2), and insert x and y, preserving the original order, at the beginning of q.

When p becomes empty, q will be a permutation of (1,2,…,N).

Find the lexicographically smallest permutation that can be obtained as q.

Constraints

N is an even number.
2≤N≤2×105
p is a permutation of (1,2,…,N).

Input

Input is given from Standard Input in the following format:

N

p1 p2 … pN

Output

Print the lexicographically smallest permutation, with spaces in between.

Sample Input 1

Copy

4

3 2 4 1

Sample Output 1

Copy

3 1 2 4

The solution above is obtained as follows:

`p`	`q`
`(3,2,4,1)`	`()`
↓	↓
`(3,1)`	`(2,4)`
↓	↓
`()`	`(3,1,2,4)`

Sample Input 2

Copy

2

1 2

Sample Output 2

Copy

1 2

Sample Input 3

Copy

8

4 6 3 2 8 5 7 1

Sample Output 3

Copy

3 1 2 7 4 6 8 5

The solution above is obtained as follows:

`p`	`q`
`(4,6,3,2,8,5,7,1)`	`()`
↓	↓
`(4,6,3,2,7,1)`	`(8,5)`
↓	↓
`(3,2,7,1)`	`(4,6,8,5)`
↓	↓
`(3,1)`	`(2,7,4,6,8,5)`
↓	↓
`()`	`(3,1,2,7,4,6,8,5)`

每次取两个数，然后把这两个相邻数放到q里，使生成的q字典序最小

这个涉及到区间查询，可以用RMQor线段树。但是那两个数要怎么维护呢，用优先队列？

然后怎么做呢？卿学姐代码做法%一波