Codeforces Round #558 (Div. 2) B2. Cat Party (Hard Edition)

Meaning of the questions: n give you the number, find the greatest x, so that makes removing one of the first x, so that the same number of other figures have emerged;

Defines two map <int, int> a, b;

a number of times to deposit entered (set to x), b for a number of the current number of times (set y) appears deposit;

Description All numbers are equal when the front i to position i, i if and only if when x * y == i! = Ans = i + 1 when n-; (due to the removal of one)

Or when x * y == i-1, i to be described when the position in front of one, i.e., meet the requirements ans = i;

 

#include <bits/stdc++.h>
using namespace std;
map<int,int>a,b;
int main()
{
    int n;
    cin>>n;
    int x;
    int ans=1;
    for(int i=1;i<=n;i++){
        cin>>x;
        a[x]++;
        b[a[x]]++;
        if(a[x]*b[a[x]]==i&&i!=n){
            ans=i+1;
        }
        if(a[x]*b[a[x]]==i-1){
            ans=i;
        }
    }
    cout<<ans<<endl;
 
    return 0;
}

View Code