牛客第四次多校Maximum Mode

链接：https://www.nowcoder.com/acm/contest/142/G
来源：牛客网

题目描述

The mode of an integer sequence is the value that appears most often. Chiaki has n integers a₁,a₂,...,a_n. She woud like to delete exactly m of them such that: the rest integers have only one mode and the mode is maximum.

输入描述:

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains two integers n and m (1 ≤ n ≤ 10^5, 0 ≤ m < n) -- the length of the sequence and the number of integers to delete.

The second line contains n integers a1,a2, ..., an(1 ≤ ai ≤ 10^9)denoting the sequence.It is guaranteed that the sum of all n does not exceed 10^6

输出描述:

For each test case, output an integer denoting the only maximum mode, or -1 if Chiaki cannot achieve it.

示例1

输入

5
5 0
2 2 3 3 4
5 1
2 2 3 3 4
5 2
2 2 3 3 4
5 3
2 2 3 3 4
5 4
2 2 3 3 4

输出

-1
3
3
3
4

#include<bits/stdc++.h>
using namespace std;
const int MAXN=1e5+10;
int a[MAXN];
map<int, int>ma;
map<int ,int>mb;
int sum[MAXN],cnt[MAXN];
int main()
{
    int _;
    scanf("%d",&_);
    while(_--)
    {
        ma.clear();
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            ma[a[i]]++;
            sum[i]=0;
            cnt[i]=0;
        }
        for(auto &i:ma) cnt[i.second]++;
        for (int i = n-1; i ; i--) {   //求出后缀和
            cnt[i]+=cnt[i+1];
            sum[i]=sum[i+1]+cnt[i];//从后向前进行，如果一个数出现n次，在向前加的过程中就会加n次。保证后缀和正确
        }
        int MAX=-1;
        for(auto &i:ma){
            if(sum[i.second]-1<=m) MAX=max(MAX,i.first);//枚举答案，-1是因为只要保证当前这个众数比其他的多一即可
        }
        printf("%d
",MAX);
    }
    return 0;
}