Polycarp's Pockets（思维）

Polycarp has nn coins, the value of the ii-th coin is aiai. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.

For example, if Polycarp has got six coins represented as an array a=[1,2,4,3,3,2]a=[1,2,4,3,3,2], he can distribute the coins into two pockets as follows: [1,2,3],[2,3,4][1,2,3],[2,3,4].

Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.

Input

The first line of the input contains one integer nn (1≤n≤1001≤n≤100) — the number of coins.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100) — values of coins.

Output

Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.

Examples

Input

6

1 2 4 3 3 2

Output

Input

1

100

Output

题解：找出现重复次数最多的元素的数量即可

AC代码1：

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<cstring>

using namespace std;

int main() {

	int n;

	cin>>n;

	int a[105];

	for(int t=0; t<n; t++) {

		scanf("%d",&a[t]);

	}

	sort(a,a+n);

	int maxn=1;

	int sum=1;

	for(int t=0; t<n-1; t++) {

		if(a[t]==a[t+1]) {

			sum++;

		} else if(a[t]!=a[t+1]) {

			sum=1;

		}

		if(maxn<sum) {

			maxn=sum;

		}

	}

	cout<<maxn<<endl;

	return 0;

}

AC代码2：

#include<cstdio>

#include<iostream>

#include<cstring>

#include<algorithm>

using namespace std;

int main()

{

	int sum[101]={0};

	int n;

	cin>>n;

	int k;

	for(int t=0;t<n;t++)

	{

		scanf("%d",&k);

		sum[k]++;

	}

	int maxn=1;

	for(int t=1;t<=100;t++)

	{

		if(sum[t]>maxn)

		{

			maxn=sum[t];

		}

	}

	cout<<maxn<<endl;

	return 0;

}

巴特西

Polycarp's Pockets（思维）

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