Codeforces #260 (Div. 2)　C. Boredom(动态规划)

C. Boredom

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
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.

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 test(s)

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

Note

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.

一道DP题，自己写的总是跪在12个样例

12个样例的测试数据是100,000,导致我差错也差不了

我的想法是用a[i]存储i出现的次数与i的乘积

因为每相邻两项总要取其中一项,所以有a[i] += max(a[i-2], a[i-3])

比如:1 2 3 4 5 6

可以选取:

1 3 5

1 3 6

1 4 6

2 4 6

循环遍历如果有a[i]!=0,则取a[i] += max(a[i-2], a[i-3]);

结果错了,想不通怎么回事

先附上正确的答案吧:

#include <cstdio>
#include <iostream>
#define MAXN 100100
#define LL long long
using namespace std;

LL n, k, a[MAXN];

int main(void) {
cin >> n;
for(LL i=0; i<n; ++i) {
cin >> k;
a[k]++;
}
//a[i]表示前i个回合得分最大值
for(LL i=2; i<100001; ++i) {
a[i] = max(a[i-2]+i*a[i], a[i-1]);//要或不要第i-1项
//cout << "a[" << i << "] = " << a[i] << endl;
}
cout << a[100000] << endl;
return 0;
}