Polycarp's Pockets (cf 1003A)

解决分配不同价值的硬币到两个口袋中,确保每个口袋内没有相同价值的硬币,以找到最少需要的口袋数。通过输入硬币的数量和价值,输出所需的最少口袋数。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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

2

input

1
100

output

1

 

代码如下:

#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <ctime>
#define maxn 10007
#define N 107
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define eps 0.000000001
#define mod 1000000007
using namespace std;
typedef long long ll;

int main()
{
    int n;
    cin>>n;
    int a[111];
    for(int i=0; i<n; i++)
    {
        int m;
        cin>>m;
        a[m]++;
    }
    int max1=-INF;
    for(int i=1; i<=100; i++)
        max1=max(max1,a[i]);
    cout<<max1<<endl;
    return 0;
}

 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值