1 second
256 megabytes
standard input
standard output
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.
给出n个数，每次能够选择消除一个值为ai的数，那么全部ai-1。ai+1的数也会被消掉，同一时候会获得值ai，问最多能够获得多少？
看完题就可想到这应该是一个dp问题，首先，哈希一下，存下每一个数的个数放在p中，消除一个数i,会获得p[i]*i的值（由于能够消除p[i]次），假设从0的位置開始向右消去，那么，消除数i时，i-1可能选择了消除，也可能没有。假设消除了i-1，那么i值就已经不存在，dp[i] = dp[i-1]。假设没有被消除。那么dp[i] = dp[i-2]+ p[i]*i。
那么初始的dp[0] = 0 ; dp[1] = p[1] ;得到了公式 dp[i] = max（dp[i-1],dp[i-2]+p[i]*i）.
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL __int64
using namespace std;
LL p[110000] , dp[110000] ;
int main()
{
LL i , n , x , maxn = -1;
memset(p,0,sizeof(p));
memset(dp,0,sizeof(dp));
scanf("%I64d", &n);
for(i = 1 ; i <= n ; i++)
{
scanf("%I64d", &x);
if(x > maxn)
maxn = x ;
p[x]++ ;
}
dp[1] = p[1] ;
for(i = 2 ; i <= maxn ; i++)
{
dp[i] = max( dp[i-1],dp[i-2]+p[i]*i );
}
printf("%I64d\n", dp[maxn]);
return 0;
}