Title Link: non zero segment division

Title Description

A1,A2,⋯,An   It's a   n   An array of natural numbers (non negative integers). We call it   Ai,⋯,Aj   Is a non-zero segment if and only if the following conditions are met at the same time:

• 1≤i≤j≤n；
• For any integer   k. If   i ≤ K ≤ j, then   Ak>0；
• i=1   or   Ai−1=0；
• j=n   or   Aj+1=0.

Here are some simple examples:

• A=[3,1,2,0,0,2,0,4,5,0,2]   Medium   four   The non-zero segments are   [3,1,2],[2],[4,5]   and   [2]；
• A=[2,3,1,4,5]   have only   one   A non-zero segment;
• A=[0,0,0]   Then there are no non-zero segments (i.e. the number of non-zero segments is)   0).

Now we can compare arrays   A   Do the following: select any positive integer   p. Then   A   All less than   p   All the numbers become   0 Try to choose a suitable one   p. Make array   A   The number of non-zero segments in reaches the maximum. If entered   A   The number of non-zero segments has reached the maximum, which can be taken   p=1, i.e. No   A   Make any changes.

Input format

Read in data from standard input.

The first line of input contains a positive integer   n.

The second line of input contains   n   A natural number separated by spaces   A1,A2,⋯,An.

Output format

Output to standard output.

Only one integer is output to represent the array   A   The maximum number of non-zero segments after operation.

Example 1 input

11
3 1 2 0 0 2 0 4 5 0 2

Sample 1 output

5

Example 1 explanation

p=2   When, A=[3,0,2,0,0,2,0,4,5,0,2], 5   The non-zero segments are   [3],[2],[2],[4,5]   and   [2]； At this time, the number of non-zero segments reaches the maximum.

Example 2 Input

14
5 1 20 10 10 10 10 15 10 20 1 5 10 15

Example 2 output

4

Data

Example 2 explanation

p=12   When, a = [0,0,20,0,0,0,15,0,20,0,0,0,0,15], 4   The non-zero segments are   [20],[15],[20]   and   [15]； At this time, the number of non-zero segments reaches the maximum.

Example 3 input

3
1 0 0

Data

Example 3 output

1

Data

Example 3 explanation

p=1   When, A=[1,0,0], only   one   Non zero segments   [1] , the number of non-zero segments reaches the maximum.

Example 4 input

3
0 0 0

Example 4 output

0

Example 4 explanation

whether   p   What is the value, A   None of them contain non-zero segments, so the number of non-zero segments is at most   0

Subtask

70%   The test data meet the requirements   n≤1000；

All test data meet   n≤5 × 10 ^ 5, and array   A   No more than   10^4.

The first version of no brain violence code... Has 70 points 2333

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
    int n;
    cin>>n;

    int* a=(int*)malloc(sizeof(int)*n);
    int* b=(int*)malloc(sizeof(int)*n);
    int* c=(int*)malloc(sizeof(int)*n);
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        b[i]=a[i];
        c[i]=a[i];
    }
    sort(b,b+n);
    
    int p=0,num=0,max=-1;
    for(int i=0;i<n;i++)
    {
        p=b[i];
        while(p==b[i])
        {
            i++;
        }
        i--;

        for(int j=0;j<n;j++)
        {
            if(c[j]<p)
            {
                c[j]=0;
            }
        }

        num=0;
        for(int j=0;j<n;j++)
        {
            if(c[j]!=0)
            {
                num++;
                while(c[j]!=0)
                {
                    j++;
                }
                j--;
            }
        }
        if(num>max)
        {
            max=num;
        }

        for(int j=0;j<n;j++)
        {
            c[j]=a[j];
        }
    }
    if(max==-1)
    {
        max=0;
    }
    cout<<max;

    return 0;
}

  After you have changed everything you can find that can be optimized... Still timeout, or 70 points

#include <iostream>
#include <cstdio>
#include <set>
using namespace std;

int main()
{
    int n;
    scanf("%d",&n);

    int* a=(int*)malloc(sizeof(int)*n);
    set<int> b;
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        b.insert(a[i]);
    }
    
    int p=0,num=0,max=0;
    for(set<int>::iterator it=b.begin();it!=b.end();it++)
    {
		p=*it;
        if(p==0)
        {
            continue;
        }
        num=0;
        for(int j=0;j<n;j++)
        {
            if(a[j]<p)
            {
                a[j]=0;
            }
            if(a[j]>=p)
            {
                num++;
                while(a[j]>=p)
                {
                    j++;
                }
                j--;
            }
        }
        if(num>max)
        {
            max=num;
        }
    }
    printf("%d",max);

    return 0;
}

It is OK! As Xiaobai, I learned new knowledge! Take notes

A friendly introduction to Xiaobai:

(23 messages) basic introduction to STL set_ Zheng Dandan's blog - CSDN blog

(23 messages) detailed usage of set in STL_ Contentment blog - CSDN blog

About accessing the set element:

(23 messages) access to elements in the set container_ Running snail CSDN blog_ Access set element

(23 messages) set in STL --- can the elements in set be modified directly_ Cognitive action persistence - CSDN blog

//Access set element
for(set<int>::iterator it=st.begin();it!=st.end();it++)
{
		cout<<*it<<" ";
}

Finally, I learned the solution of the boss!

(23 messages) csp2021-09-2 non-zero segment division_ Next door Li sou's blog - CSDN blog

This metaphor is really good! Make the meaning clear at once √

Although the reflection arc is long, I still stare at the code for a long time to reflect how it is implemented, and I may not write it next time

Notes notes

unique() function

#include <algorithm>

This is a more specific introduction:

(23 messages) unique function in C + +_ Simao CSDN blog

(23 messages) explanation of unique() function_ Bearded blog - CSDN blog_ unique

[finishing] unique function in C + + - nimphy - blog Garden (cnblogs.com)

(little voice, take a good look at the interface of this blog. After opening it, I played with the villain in the lower right corner for a long time before I thought of coming to see and explain XD