Educational Codeforces Round 34 C. Boxes Packing【模拟/STL-map/俄罗斯套娃】

C. Boxes Packing

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mishka has got n empty boxes. For every i (1 ≤ i ≤ n), i-th box is a cube with side length ai.

Mishka can put a box i into another box j if the following conditions are met:

• i-th box is not 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 calledvisible iff it is not 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 has 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 is possible to put box 1 into box 2, and 2 into 3.

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

【分析】：之前没用map而是hash就一直RE···当然思路就是找出出现次数最多的数的次数直接输出。

【代码】：

#include <iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<streambuf>
#include<cmath>
#include<string>
using namespace std;
#define ll long long
#define oo 10000000
const int N = +;
int a,m[N];
int ans;
/*
直接排序找出出现次数最多的那个数的次数直接输出
*/
int main()
{
    int n;
    int ma=-;
    memset(m,,sizeof(m));
    scanf("%d",&n);
    for(int i=;i<n;i++)
    {
        scanf("%d",&a);
        m[a]++;
    }
    //sort(m,m+5000);
    for(int i=;i<+;i++)
    {
        if(m[i]>ma)
            ma=m[i];
    }
    printf("%d\n",ma);
    return ;
}

RE代码

#include <bits/stdc++.h>
using namespace std;
int n, x, ma;
map<int,int> m;
int main()
{
    cin >> n;
    for (int i = ; i < n; i++)
    {
        cin >> x;
        m[x]++;
        ma = max(ma, m[x]);
    }
    printf("%d\n", ma);
}

AC代码