链接:https://www.nowcoder.com/acm/contest/142/G
来源:牛客网
时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 131072K,其他语言262144K
64bit IO Format: %lld
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
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;
#define LL long long
#define clr(a) memset(a,0,sizeof(a))
const int MAXN = 1e5+10;
const int INF = 0x3f3f3f3f;
const int N = 2010;
int n,m,t;
int a[MAXN],cnt[MAXN],sum[MAXN];
map<int,int>mp;
int main(){
scanf("%d",&t);
while(t--){
mp.clear();
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
mp[a[i]]++;
sum[i] = 0;
cnt[i] = 0;//记录众数出现的个数
}
map<int,int>::iterator it;
for(it=mp.begin();it!=mp.end();it++){
cnt[(*it).second] ++;
}
for(int i=n-1;i;i--){
cnt[i] += cnt[i+1];
sum[i] = sum[i+1]+cnt[i];
}
int maxx = -1;
for(it=mp.begin();it!=mp.end();it++){
if(sum[(*it).second]-1 <= m)
maxx = max(maxx,(*it).first);
}
printf("%d\n",maxx);
}
return 0;
}