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.
InputThe 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.
OutputPrint 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.
Examples6 1 2 4 3 3 2
2
1 100
1
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
int a[110];
int main()
{
int n,i,j,x;
cin>>n;
memset(a,0,sizeof(a));
for(i=0;i<n;i++)
{
cin>>x;
a[x]++;
}
int ans=-1;
for(i=1;i<=100;i++)
{
ans = max(ans,a[i]);
}
cout<<ans<<endl;
return 0;
}
本文介绍了一个有趣的算法问题:如何用最少数量的口袋来存放不同价值的硬币,确保相同价值的硬币不会放在同一个口袋中。通过分析输入的硬币数组,文章提供了一种高效的解决方案,并附带了实现这一目标的C++代码。
630

被折叠的 条评论
为什么被折叠?



