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69 lines
2.0 KiB
Plaintext
69 lines
2.0 KiB
Plaintext
#FAMOUS DP PROBLEM WITH EXPLANATION FOR BEGGINNERS
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A. Boredom
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time limit per test1 second
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memory limit per test256 megabytes
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inputstandard input
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outputstandard output
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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.
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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.
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Alex is a perfectionist, so he decided to get as many points as possible. Help him.
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Input
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The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex's sequence.
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The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105).
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Output
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Print a single integer — the maximum number of points that Alex can earn.
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Examples
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inputCopy
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2
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1 2
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outputCopy
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2
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inputCopy
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3
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1 2 3
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outputCopy
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4
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inputCopy
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9
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1 2 1 3 2 2 2 2 3
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outputCopy
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10
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Note
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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.
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SOLUTION :-
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#include<bits/stdc++.h>
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using namespace std;
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int main()
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{
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int n,x;
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cin>>n;
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long long int a[100005]={0};
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long long int dp[100005]={0};
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for(int i=0;i<n;i++)
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{
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cin>>x;
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a[x]++;// cREATING A FREQUENCY ARRAY USING ITS INDEX AS NUMBERS
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}
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dp[0]=0; //if INPUT ARRAY ONLY HAS 0
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dp[1]=a[1];//if input ARRAY ONLY HAS 0 AND 1'S
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for(int i=2;i<=100000;i++)
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{
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dp[i]=max(dp[i-1],dp[i-2]+(i*a[i]));//IF NOT TAKING ELMENT THEN TAKE THE PREVIOUS ONE
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//ELSE TAKE THE PREVIOUS TO PREVIOUS AND ADD THE FREQ OF PRESENT ELEMENT
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}
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/* for(int i=0;i<=1000;i++)
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{
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cout<<dp[i]<<" ";
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}
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cout<<endl;*/
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cout<< dp[100000]<<endl;
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}
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