思路

hash表存val值的个数t，如果 t能被 val+1 整除，sum加上t；如果不能被整除，加上(t/(val+1)+1)*(val+1)

class Solution {
public:
    int numRabbits(vector<int>& answers) {
        map<int,int> mp;
        for(auto it : answers){
            mp[it] ++ ;
        }
        int sum = 0 ;
        for(auto it : mp){
            int val = it.first, t = it.second;
            sum+=((t-1)/(val+1)+1)*(val+1);
        }
        return sum;
    }
};