题目链接：
A. Polycarp's Pockets
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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
Copy
6 1 2 4 3 3 2
output
Copy
2
input
Copy
1 100
output
Copy
1
代码实现：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN=5010;
const int MAXM=100010;
const int M=50010;
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int a[110],vis[150];
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
vis[a[i]]++;
}
int maxx=-10;
for(int i=1;i<150;i++)
maxx=max(maxx,vis[i]);
cout<<maxx<<endl;
}
return 0;
}
