dp解Codeforces Round #260 (Div. 2)C. Boredom

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
long long shu[100010],fu[100010];
int main()
{
	int t,n,i,r=-1,l=100010;
	cin>>n;
	for(i=0;i<n;i++)
	{
		cin>>t;
		shu[t]++;
		l=min(l,t);
		r=max(r,t);
	}
	fu[1]=1*shu[1];
	for(i=max(l,2);i<=r;i++)
		fu[i]=max(fu[i-1],fu[i-2]+shu[i]*i);
	//前i个数(包括i)所能得到的最大分数等于,前i-1个数所能得到的最大分数,和选择第i个数所能得到的最大分数
	//它们俩的最大值
	cout<<fu[r];
}

dp解Codeforces Round #260 (Div. 2)C. Boredom

时间: 2024-12-26 01:03:53

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