codeforces 557 C

由于期末。非常久没刷题了，CF一直掉……

这个题事实上非常简单。

。由于做法非常easy想到嘛。。

就是枚举max=x时，最大能保留多少价值。不断更新ans,

结果就是全部价值和减去ans就好

因为最大可以保留的长度是199+200，所以当max=x时。算最大能保留多少价值，

也是一个循环算出当前长度比x小的那个桌子角的最大的那几个价值之和保留即可了，

这里写的比較绕。。反正看看代码一下就懂了。。。

维护一个map即可了

比赛的时候，由于好久没刷题了。一直没想清楚，一直到比赛结束前还剩下十分钟才恍然大悟。

。

复杂度是（logn+200）*n

只是也没办法啦。要期末嘛

C. Arthur and Table

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Arthur has bought a beautiful big table into his new flat. When he came home, Arthur noticed that the new table is unstable.

In total the table Arthur bought has n legs, the length of thei-th leg isl_i.

Arthur decided to make the table stable and remove some legs. For each of them Arthur determined numberd_i — the amount of energy that he spends to remove thei-th leg.

A table with k legs is assumed to be stable if there are more than half legs of the maximum length. For example, to make a table with5 legs stable, you need to make sure it has at least three (out of these five) legs of the maximum length. Also, a table with one leg is always stable and a table with two legs is stable if and only if they have the same lengths.

Your task is to help Arthur and count the minimum number of energy units Arthur should spend on making the table stable.

Input

The first line of the input contains integer n (1 ≤ n ≤ 10⁵) — the initial number of legs in the table Arthur bought.

The second line of the input contains a sequence of n integersl_i (1 ≤ l_i ≤ 10⁵), wherel_i is equal to the length of thei-th leg of the table.

The third line of the input contains a sequence of n integersd_i (1 ≤ d_i ≤ 200), whered_i is the number of energy units that Arthur spends on removing thei-th leg off the table.

Output

Print a single integer — the minimum number of energy units that Arthur needs to spend in order to make the table stable.

Sample test(s)

Input

2
1 5
3 2

Output

2

Input

3
2 4 4
1 1 1

Output

0

Input

6
2 2 1 1 3 3
4 3 5 5 2 1

Output

8

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
map<int,int>num;
struct Box
{
	int len,val;
}box[100010];
bool cmp(Box one,Box two)
{
	return one.len<two.len;
}
int main()
{
	int n;
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>box[i].len;
	for(int i=0;i<n;i++)
		cin>>box[i].val;
	sort(box,box+n,cmp);
	int l=-1,r=-2,ans=-1;
	for(int i=0;i<n;i++)
		for(int j=i;j<n;j++)
			if(box[j+1].len!=box[i].len)
			{
				for(int k=l;k<=r;k++)
					num[box[k].val]++;
				l=i;r=j;
				map<int,int>::iterator it=num.end();
				int len=j-i,sum=0;
				while(len>0)
				{
					if(it==num.begin())
						break;
					it--;
					for(int k=0;k<it->second&&len>0;k++,len--)
						sum+=it->first;
				}
				for(int k=l;k<=r;k++)
					sum+=box[k].val;
				ans=max(ans,sum);
				i=j;
				break;
			}
	int sum=0;
	for(int i=0;i<n;i++)
		sum+=box[i].val;
	cout<<sum-ans;
}