codeforce 455A—— DP—— Boredom
2152 단어 code
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;
}
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