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 a1,a2,...,an 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≤105,0≤m<n) ( 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≤109) 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 106 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 n u m 的个数，再给这些个数 count c o u n t 从小大排序，从 num n u m 最大的开始验证，验证是如果答案是 numi n u m i ,那么个数设为 counti c o u n t i ,那么所以比 counti c o u n t i 大的值都必须降低到 counti−1 c o u n t 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
#include
#include
#include
#include
#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]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;iscanf("%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//离散化
        for(int i=0;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: "<
                ans=id[i];
                sign=true;
                break;
            }
        }
        if(sign)printf("%d\n",ans);
        else printf("-1\n");
    }
    return 0;
}