Maximum Mode【Nowcoder多校训练第4场G题】（离散化+二分）

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$a_1,a_2,...,a_n$. 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 $a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9)$ 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
我的做法好像有点复杂啊。先离散化，然后，统计每个离散化后出现的数字$num$的个数，再给这些个数$count$从小大排序，从$num$最大的开始验证，验证是如果答案是$num_i$,那么个数设为$count_i$,那么所以比$count_i$大的值都必须降低到$count_i-1$，这是最低值，那么我们就可以验证了，这里就是二分在加上前缀和了
代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<map>
#define maxx 100005
using namespace std;
int n,m;
int a[maxx];
int id[maxx];
int cnt,num;
int get(int x)
{
    int l=1,r=num;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(id[mid]==x) return mid;
        if(id[mid]<x) l=mid+1;
        else r=mid-1;
    }
    return 0;
}
int b[maxx];
int order[maxx];
int sum[maxx];
int tot;
int _get(int x)
{
    int l=1,r=tot;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(order[mid]>=x) r=mid-1;
        else l=mid+1;
    }
    return l;
}
void init()
{
    cnt=0;
    tot=0;
    memset(b,0,sizeof(b));
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        init();
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",a+i),id[++cnt]=a[i];
        num=unique(id+1,id+cnt+1)-id-1;
        sort(id+1,id+num+1);
        for(int i=0;i<n;i++) a[i]=get(a[i]);//离散化
        for(int i=0;i<n;i++) b[a[i]]++;//统计个数
        for(int i=1;i<=num;i++)
            if(b[i])order[++tot]=b[i];//把个数收集起来
        sort(order+1,order+1+tot);
        for(int i=1;i<=tot;i++)
            sum[i]=sum[i-1]+order[i];//前缀和
        bool sign=false;
        int ans;
        for(int i=num;i>0;i--)
        {
            int index=_get(b[i]);//二分去找边界，就是找到所有大于等于b[i]的最小值
            int total=sum[tot]-sum[index];
            if(total-(b[i]-1)*(tot-index)<=m)//验证
            {
                //cout<<"xixi: "<<total-(b[i]-1)*(tot-index)<<endl;
                ans=id[i];
                sign=true;
                break;
            }
        }
        if(sign)printf("%d\n",ans);
        else printf("-1\n");
    }
    return 0;
}