牛客多校训练营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,...,an. 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 ≤ 105, 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 ≤ 109) denoting the sequence.
It is guaranteed that the sum of all n does not exceed 106.

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

复制

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

复制

-1
3
3
3
4

题意：每次删去m个数找出删除后尽量众数最多并且最大的一个数。

#include<bits/stdc++.h>//n,m<=1e5 ,a[i]<=1e9
using namespace std;
const int N=100005;
typedef struct mode{
    int x,num;
    bool operator < (const mode & p1) const {
        return num<p1.num;
    }
}mode;
int cmp(int a,int b){
    return a>b;
}
int main()
{
    int n,m;
    int T;
    scanf("%d",&T);
    while(T--){
        int sum[N]={0};
        int a[N]={0};
        mode b[N];
        map<int,int>mp;
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            mp[a[i]]++;
        }
        int cnt=0;
        map<int,int>::iterator it=mp.begin();
        for(;it!=mp.end();it++){
            b[cnt].x=it->first;
            b[cnt].num=it->second;
            cnt++;
        }
        sort(b,b+cnt);
        for(int i=cnt-1;i>=0;i--){
            sum[i]=sum[i+1]+b[i].num;
        }
        /*for(int i=0;i<cnt;i++){
            printf("sum[%d] : %d\n",i,sum[i]);
        }*/
        map<int,int>mp2;
        //printf("MMMMMMMMMMM:%d\n",m);
        int maxz=-1;
        for(int i=0;i<cnt;i++){
            int t=mp2[b[i].num]*b[i].num;
            //printf("mp2[b[i].x] :%d  b[i].num: %d        t: %d \n",mp2[b[i].num],b[i].num,t);
            int tt=t+sum[i+1]-(mp2[b[i].num]+cnt-i-1)*(b[i].num-1);
            //printf("SUM[i+1] %d  ,      cnt-i-1:     %d\n",sum[i+1],cnt-i-1);
            //printf("%d\n",tt);
            if(tt<=m){
                maxz=max(b[i].x,maxz);
            }
            mp2[b[i].num]++;
        }
        printf("%d\n",maxz);
    }
    return 0;
}