Maximum Mode

链接：https://www.nowcoder.com/acm/contest/142/G
来源：牛客网

Maximum Mode

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

复制

输出

复制

题意：mode 数是出现次数最多的数。

输入n个数，删除m个数后mode数只能有一个，并且要尽量的大。

方法：记录每个数出现的次数，枚举每个数作为mode数时至少需要删除其他多少数。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
map<ll,ll>mp;
struct node
{int val,num,sum;
};
node b[maxn];
int a[maxn];
bool cmp(node a,node b)
{return a.num<b.num;
}
int main()
{int t,n,m,i,cnt,k,sum2,sum1;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(i=1;i<=n;i++)scanf("%d",&a[i]);sort(a+1,a+1+n);cnt=0;k=0;sum2=0;for(i=1;i<n;i++){if(a[i]==a[i+1]){cnt++;}else{b[++k].val=a[i];b[k].num=cnt+1;b[k].sum=-1;sum2+=b[k].num;cnt=0;}}if(a[n-1]==a[n]){b[++k].val=a[n];b[k].num=cnt+1;b[k].sum=-1;sum2+=b[k].num;}else{b[++k].val=a[n];b[k].num=1;b[k].sum=-1;sum2+=1;}sort(b+1,b+1+k,cmp);ll temp;cnt=2;//cout<<sum2<<endl;b[1].sum=sum2-b[1].num-(b[1].num-1)*(k-1);//cout<<b[1].sum<<"!!"<<endl;sum1=b[1].num;for(i=2;i<=k;i++){if(b[i].num==b[i-1].num){b[i].sum=b[i-1].sum;cnt++;sum1+=b[i].num;}else{temp=sum2-sum1-b[i].num-(b[i].num-1)*(k-cnt);b[i].sum=temp;cnt++;sum1+=b[i].num;}}int maxn=-1;int flag=0;for(i=1;i<=k;i++){if(m>=b[i].sum){flag=1;maxn=max(maxn,b[i].val);}}if(flag==0)printf("-1\n");elseprintf("%d\n",maxn);}
}
/*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*/
/*1
18 15
2 2 9 9 1 1 3 3 3 4 4 4 5 5 5 5 5 5*/

Maximum Mode相关推荐

【C++】C++11 STL算法(六)：最小/最大操作(Minimum/maximum operations)、比较运算(Comparison operations)
目录最小/最大操作(Minimum/maximum operations) 一.max 1.原型: 2.说明: 3.官方demo 二.max_element 1.原型: 2.说明: 3.官方demo ...
Lintcode42 Maximum Subarray II solution 题解
[题目描述] Given an array of integers, find two non-overlapping subarrays which have the largest sum.The ...
[LintCode] Maximum Subarray 最大子数组
Given an array of integers, find a contiguous subarray which has the largest sum. Notice The subarra ...
UVA11059 Maximum Product
问题链接:UVA11059 Maximum Product.基础级练习题,用C语言编写程序. 题意简述:输入n个整数序列,有正有负,求这个序列中最大连续累乘的子序列,其最大的值为多少.如果结果为负数, ...
Leetcode | Binary Tree Maximum Path Sum
Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. ...
贪心 ---- E. Maximum Subsequence Value[位运算]
E. Maximum Subsequence Value 题目大意:有点难解释..建议自己看题.我这里就粗略解释:给定一个数组aaa,要求选出具有最大价值的子序列.假设此子序列的长度为kkk,那么最大 ...
Codeforces Round #665 (Div. 2) Maximum Distributed Tree（树上贪心）
整理的算法模板合集: ACM模板点我看算法全家桶系列!!! 实际上是一个全新的精炼模板整合计划 CF1401D Maximum Distributed Tree(树上贪心) 给定一棵 nnn 个节点 ...
Codeforces 454C - Little Pony and Expected Maximum
454C - Little Pony and Expected Maximum 思路: m面的骰子掷n次,总共有m^n种情况,如果一种情况的最大值是m,那么它肯定包含m,那我们在所有情况下挖掉不包含m ...
pandas使用max函数和min函数计算dataframe日期（时间）数据列中最大日期和最小日期对应的数据行(maximum and minimum date or time row)
pandas使用max函数和min函数计算dataframe日期(时间)数据列中最大日期和最小日期对应的数据行(maximum and minimum date or time row in data ...
pandas计算滑动窗口中的最大值实战（Rolling Maximum in a Pandas Column）：计算单数据列滑动窗口中的最大值、计算多数据列滑动窗口中的最大值
pandas计算滑动窗口中的最大值实战(Rolling Maximum in a Pandas Column):计算单数据列滑动窗口中的最大值.计算多数据列滑动窗口中的最大值目录

Maximum Mode

题目描述

输入描述:

输出描述:

输入

输出

Maximum Mode相关推荐

最新文章

热门文章