牛客网暑期ACM多校训练营（第四场）G(Maximum Mode)

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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.

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

题意：给出n个数，m次操作，每次操作可以删除一个数，问删除m个数后剩下的数的众数最大可以是多少。

思路：因为想让一种数成为众数必须将比它出现次数多的数全部减少的比该数少，那么先处理出，出现各个次数的数的种类，变更新边找出最大值，然后从大到小枚举出现次数的容器，然后前缀和表示依次削取前面各种数一个，递推下来找到满足条件的最大值就可以了。

队友代码：

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
inline ll read()
{
register ll c=getchar(),fg=1,sum=0;
while(c>'9'||c<'0') {
if(c == '-')fg = -1;c=getchar();}
while(c<='9'&&c>='0'){sum=sum*10+c-'0';c=getchar();}
return fg*sum;
}
const int maxn=100010;
ll n,m,k;
ll a[maxn];
ll c[maxn];
vector<ll>vc[maxn];
ll ans,ct,cnt,tmp,flag;
map<ll,ll>mp;
void init(ll n)
{
mp.clear();
for(ll i=0;i<=n+5;i++)
{
vc[i].clear();
c[i]=0;
}
}
int main()
{
ll T;
T=read();
while(T--)
{
n=read();m=read();
init(n);
for(ll i=1;i<=n;i++)
{
a[i]=read();
if(mp.count(a[i])==0)mp[a[i]]=1;
else mp[a[i]]++;
}
for(ll i=1;i<=n;i++)
{
ll x=mp[a[i]];
if(x) {
vc[x].push_back(a[i]);
c[x]=max(c[x],a[i]);
mp[a[i]]=0;
}
}
ll ma=-1;
ll pre=0;
ans=-1;
for(ll i=n;i>0;i--)
{
ll xx=vc[i].size();
pre+=xx;
if(xx){
ma=max(ma,c[i]);
if(pre-1<=m)
{
ans=max(ans,ma);
}
else break;
}
m-=pre;
if(m<0) break;
}
printf("%lld\n",ans);
}
return 0;
}