rsa Round #71 (Div. 2 only)

Replace A

Time limit: 1000 ms
Memory limit: 256 MB

 

You are given a string SS containing only letters A or B. You can take any two adjacent As and replace them by a single A. You perform operations as long as possible. Print the final string.

Standard input

The first line contains the string SS.

Standard output

Print the final string on the first line.

Constraints and notes

• SS contains between 11 and 100100 characters.

Input	Output	Explanation
AAABB	ABB	At the first step you can choose the first two As, obtaining AABB. At the second step, you can choose the only two adjacent As remaining, obtaining ABB. You cannot do any more operations on this string.

和上次CF的A一样，直接暴力

#include<bits/stdc++.h>
using namespace std;
const int N=2005;
int main()
{
    ios::sync_with_stdio(false);
    string s,t;
    cin>>s;
    set<char>S;
    S.insert('A');
    while(true)
    {
        int f=1;
        for(int i=1;s[i]&&f;i++)
        {
            if(S.count(s[i-1])&&S.count(s[i]))
            {
                f=0;
                t=s.substr(0,i)+s.substr(i+1);
            }
        }
        if(f)break;
        s=t;
    }
    cout<<s;
    return 0;
}

Matrix Balls

Time limit: 1000 ms
Memory limit: 256 MB

 

You are given a matrix AA of size N \times MN×M, containing distinct elements. Initially there is a ball placed in every cell of the matrix. Each ball follows the following movement:

• If the current cell is smaller than all of its (at most 88) neighbours, the ball stops in this cell;
• otherwise, the ball moves to the smallest neighbouring cell.

Find for each cell of AA how many balls will end up there, when all the balls stop moving.

Standard input

The first line contains two integers NN and MM.

Each of the next NN lines contains MM integers representing the elements of AA.

Standard output

Print NN lines, each containing MM integers. The j^{th}j​th​​ element on the i^{th}i​th​​ line should represent the number of balls that end up in cell (i, j)(i,j).

Constraints and notes

• 1 \leq N, M \leq 5001≤N,M≤500 
• 0 \leq A_{i, j} \leq 3*10^50≤A​i,j​​≤3∗10​5​​

Input	Output	Explanation
3 3 1 3 4 5 6 7 8 9 2	6 0 0 0 0 0 0 0 3	Considering a ball in each cell, matrix[i][j]matrix[i][j] represents the value to which that ball will move 00 represents a ball that does not move to any other location (the cell is the smallest of the 88 neighbours) 1 2 3 4     0 1 3 1 1 2 5 2 0           The value on which the ball will be placed after it moved as described in the statement 1 2 3 4     1 1 1 1 1 2 1 2 2
1 6 10 20 3 4 5 6	1 0 5 0 0 0
4 4 20 2 13 1 4 11 10 35 3 12 9 7 30 40 50 5	0 4 0 4 0 0 0 0 4 0 0 0 0 0 0 4	The value on which the ball will be placed after it moved as described in the statement 1 2 3 4 5     2 2 1 1 2 2 1 1 3 3 5 5 3 3 5 5

一个球可以向八个方向滚，滚到最小的里面，问最后每个那里面放几个

必须要记忆化搜索，但是要多搜几个方向的

#include<bits/stdc++.h>
using namespace std;
const int N=505;
int a[N][N],b[N][N],n,m;
int d[8][2]= {1,0,1,1,1,-1,0,1,0,-1,-1,0,-1,1,-1,-1};
int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
            cin>>a[i][j],b[i][j]=1;
    int f=1;
    while(f)
    {
        f=0;
        for(int i=0; i<n; i++)
            for(int j=0; j<m; j++)
                if(b[i][j])
                {
                    int x=i,y=j;
                    for(int k=0; k<8; k++)
                    {
                        int tx=i+d[k][0],ty=j+d[k][1];
                        if(!(tx<0||tx>=n||ty<0||ty>=m)&&a[tx][ty]<a[x][y])
                        x=tx,y=ty;
                    }
                    if(x!=i||y!=j)
                        b[x][y]+=b[i][j],b[i][j]=0,f=1;
                }
    }
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<m; j++)
            cout<<b[i][j]<<" ";
        cout<<"\n";
    }
    return 0;
}

按照数不同往下搜索

 

#include<bits/stdc++.h>
using namespace std;
const int N=3e5+5;
int d[8][2]= {1,0,1,1,1,-1,0,1,0,-1,-1,0,-1,1,-1,-1};
int n,m,a[505][505],b[505][505];
pair<int,int>g[N];
int main()
{
    cin>>n>>m;
    for(int i=1; i<=n; i++)
        for(int j=1; j<=m; j++)
            b[i][j]=1,cin>>a[i][j],g[a[i][j]]=make_pair(i,j);
    for(int i=300000; i>=0; --i)
    {
        if(g[i].first==0)continue;
        int x=g[i].first,y=g[i].second,mi=i,sx=0,sy=0;
        for(int j=0; j<8; j++)
        {
            int tx=x+d[j][0],ty=y+d[j][1];
            if(tx<1||ty<1||tx>n||ty>m)continue;
            if(a[tx][ty]<mi)
                mi=a[tx][ty],sx=tx,sy=ty;
        }
        if(mi!=i)b[sx][sy]+=b[x][y],b[x][y]=0;
    }
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=m; j++)
            cout<<b[i][j]<<" ";
        cout<<"\n";
    }
    return 0;
}

 

Binary Differences

Time limit: 1000 ms
Memory limit: 256 MB

 

You are given a binary array AA of size NN. We define the cost of a subarray to be the number of 00s minus the number of 11s in the subarray. Find the number of distinct values KK such that there is at least one subarray of cost KK.

Standard input

The first line contains one integer NN.

The second line contains NN integers (00 or 11) representing the elements of AA.

Standard output

Print the answer on the first line.

Constraints and notes

• 1 \leq N \leq 10^51≤N≤10​5​​ 
• The subarray may be empty

Input	Output	Explanation
3 0 1 0	3	We have 33 different costs: [1, 0][1,0] has cost 00  [1][1] has cost -1−1  [0, 1, 0][0,1,0] has cost 11
4 1 0 0 1	4	[1, 0][1,0] has cost 00  [1, 0, 0][1,0,0] has cost 11  [0, 0][0,0] has cost 22  [1][1] has cost -1−1

这个是模拟

#include<bits/stdc++.h>
using namespace std;
int n,mi,ma,x,s,mis,mas;
int main()
{
    cin>>n;
    for(int i=1; i<=n; ++i)
    {
        cin>>x;
        if(!x)s++;
        else s--;
        mi=min(mi,s-mas),ma=max(ma,s-mis),mas=max(mas,s),mis=min(mis,s);
    }
    cout<<ma-mi+1;
    return 0;
}