Educational Codeforces Round 34 (Rated for Div. 2) C. Boxes Packing

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 a_i.

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 (a_i < a_j).

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 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, ..., a_n (1 ≤ a_i ≤ 10⁹), where a_i 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

 

题目大意：  小盒子可以放进比它大盒子，求这些盒子最后变成几个无法再组装的盒子

正常AC代码：

#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <string>
#include <ctype.h>
//******************************************************
#define lrt (rt*2)
#define rrt  (rt*2+1)
#define LL long long
#define inf 0x3f3f3f3f
#define pi acos(-1.0)
#define exp 1e-8
//***************************************************
#define eps             1e-8
#define inf             0x3f3f3f3f
#define INF             2e18
#define LL              long long
#define ULL             unsigned long long
#define PI              acos(-1.0)
#define pb              push_back
#define mk              make_pair

#define all(a)          a.begin(),a.end()
#define rall(a)         a.rbegin(),a.rend()
#define SQR(a)          ((a)*(a))
#define Unique(a)       sort(all(a)),a.erase(unique(all(a)),a.end())
#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(min(a,b),min(c,d))
#define max4(a,b,c,d)   max(max(a,b),max(c,d))
#define max5(a,b,c,d,e) max(max3(a,b,c),max(d,e))
#define min5(a,b,c,d,e) min(min3(a,b,c),min(d,e))
#define Iterator(a)     __typeof__(a.begin())
#define rIterator(a)    __typeof__(a.rbegin())
#define FastRead        ios_base::sync_with_stdio(0);cin.tie(0)
#define CasePrint       pc('C'); pc('a'); pc('s'); pc('e'); pc(' '); write(qq++,false); pc(':'); pc(' ')
#define vi              vector <int>
#define vL              vector <LL>
#define For(I,A,B)      for(int I = (A); I < (B); ++I)
#define FOR(I,A,B)      for(int I = (A); I <= (B); ++I)
#define rFor(I,A,B)     for(int I = (A); I >= (B); --I)
#define Rep(I,N)        For(I,0,N)
#define REP(I,N)        FOR(I,1,N)
using namespace std;
const int maxn=1e5+;
vector<int>Q[maxn];
const int MOD=1e9+;
int a[],vis[];
int k=;

int main()
{
    int n,x;
    while(~scanf("%d",&n))
    {
        memset(vis,,sizeof(vis));
        Rep(i,n) scanf("%d",&a[i]);
        sort(a,a+n);
        int ans=;
        for(int i=;i<n;i++)
        {
            if(!vis[i])
            {
                x=a[i],vis[i]=;
                for(int j=i+;j<n;j++)
                {
                    if(!vis[j])
                    {
                        if(x<a[j])
                        {
                            x=a[j],vis[j]=;
                        }
                    }
                }
                ans++;
            }
        }
         cout<<ans<<endl;
    }
    return ;
}

map巧用，贼短的代码：

#include<stdio.h>
#include<iostream>
#include<map>
#define For(I,A,B)      for(int I = (A); I < (B); ++I)
#define FOR(I,A,B)      for(int I = (A); I <= (B); ++I)
#define rFor(I,A,B)     for(int I = (A); I >= (B); --I)
#define Rep(I,N)        For(I,0,N)
#define REP(I,N)        FOR(I,1,N)
using namespace std;
#define LL long long
const int maxn=+;
LL a[maxn],sum[maxn];
map<LL,LL>b;
LL num,Max;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        b.clear();
        Max=-;
        REP(i,n)
        {
            scanf("%I64d",&num);
            b[num]++;
            Max=max(Max,b[num]);
        }
        cout<<Max<<endl;
    }
    return ;
}