Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex's sequence.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105).
Print a single integer — the maximum number of points that Alex can earn.
2
1 2
2
3
1 2 3
4
9
1 2 1 3 2 2 2 2 3
10
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this[2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.
比赛中遇到的dp题,比赛时没有思路,赛后有点思路但不完善,听了讲解后算是懂了,还需要多积累。
若取当前的值,则与其相邻的值就不能取,故状态转移方程:
dp[i][0]=max(dp[i-1][0],dp[i-1][1]);
dp[i][1]=dp[i-1][0]+value[i];
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; long long vis[100005]; long long dp[100005][2]; long long value[100005]; int main() { int n,num; scanf("%d",&n); for(int i=1; i<=n; i++) { scanf("%d",&num); vis[num]++; } int cnt=1; for(int i=1;i<=1e5;i++) { dp[i][0]=max(dp[i-1][1],dp[i-1][0]); dp[i][1]=dp[i-1][0]+vis[i]*i; } //cout<<dp[cnt-1][0]<<endl<<dp[cnt-1][1]<<endl; printf("%I64d ",dp[100000][0]>dp[100000][1]?dp[100000][0]:dp[100000][1]); return 0; }