codeforce 455A—— DP—— Boredom

2152 단어 code
Description
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 bringsak points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
Input
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).
Output
Print a single integer — the maximum number of points that Alex can earn.
Sample Input
Input
2
1 2

Output
2

Input
3
1 2 3

Output
4

Input
9
1 2 1 3 2 2 2 2 3

Output
10

Hint
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.
#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

long long  dp[100010][2];

int a[100010];

int temp[100010];

int n;

int main()

{

      while(~scanf("%d", &n)){

          memset(dp, 0, sizeof(dp));

        for(int i = 1; i <= n; i++){

            scanf("%d", &a[i]);

            temp[a[i]]++;

        }



        dp[1][1] = temp[1];

        dp[1][0] = 0;

        for(int i = 2; i <= 100000; i++){

            dp[i][1] = dp[i-1][0] + 1ll*temp[i]*i;

            dp[i][0] = max(dp[i-1][1], dp[i-1][0]);

        }



        printf("%lld
", max(dp[100000][1], dp[100000][0])); } return 0; }

좋은 웹페이지 즐겨찾기