Educational codeforces Round C. Boxes Packing

Original question:
C. Boxes Packing
Time limit per test1 second
Memory limit per test256 megabytes
Inputstandard input
Outputstandard output
Mishka has got n empty boxes. For every I (1≤i≤n), i-th Box was a cube with side length AI.

Mishka can put a box I into another box J if the following conditions is met:

I-th box is isn't put into another box;
j-th box doesn ' t contain any other boxes;
Box I is smaller than box J (Ai < AJ).
Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it isn't put into some another box.

Help Mishka to determine the minimum possible number of visible boxes!

Input
The first line contains one integer n (1≤n≤5000)-the number of boxes Mishka have got.

The second line contains n integers a1, a2, ..., an (1≤ai≤109), where AI is the side length of i-th box.

Output
Print the minimum possible number of visible boxes.

Examples
Input
3
1 2 3
Output
1
Input
4
4 2 4 3
Output
2
Note
In the first example it's possible to put box 1 into Box 2, and 2 into 3.

In the second example Mishka can put box 2 to box 3, and box 4 into box 1.

English:
Here you have n boxes, each with a size AI, and a large box to fit the small box. Now ask how many boxes you can have left in case you try to pack the boxes.

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
struct box
{
    int s;
    bool in;
    box()
    {
        s=0;
        in=false;
    }
};
box b[5001];
int cmp(const box &b1,const box &b2)
{
    return b1.s<b2.s;
}
int main()
{
    ios::sync_with_stdio(false);
    int n;
    while(cin>>n)
    {
        memset(b,0,sizeof(box));
        for(int i=1;i<=n;i++)
        {
            cin>>b[i].s;
            b[i].in=false;
        }
        sort(b+1,b+1+n,cmp);
        int cnt=n;
        for(int i=1;i<=n;i++)
        {
            int tmp=i;
            if(b[i].in)
                continue;
            for(int j=i+1;j<=n;j++)
            {
                if(b[j].s>b[tmp].s&&!b[j].in)
                {
                    b[j].in=1;
                    tmp=j;
                    cnt--;
                }
            }
        }
        cout<<cnt<<endl;
    }
    return 0;
}

Answer:

According to the box size from small to large, small boxes in a slightly larger box than it, the box is marked with a box, and finally check the number of boxes left to go.