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
a
k
) and delete it, at that all elements equal to
a
k
+ 1
and
a
k
- 1
also must be deleted from the sequence. That step brings
a
k
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
≤ 10
5
) that shows how many numbers are in Alex’s sequence.
The second line contains
n
integers
a
1
,
a
2
, …,
a
n
(
1 ≤
a
i
≤ 10
5
).
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.
/*
题目链接： http://codeforces.com/problemset/problem/456/C
题意：给定一个含有n个整数的数组，你可以进行多次操作，每次操作从数组选一个数a[k]，
然后将其删除，然后删除与a[k]-1和a[k]+1相等的数，
然后可以得到a[k]分，求进行多次操作后得到的最多的分
DP
先将每个数*这个数的个数，得到选这个数所能得到的分
然后DP,
dp[i] = max( dp[i-1]， 不选i这个数
dp[i-2] + val[i] 选i这个数 )
要注意的是，
最终的答案不是dp[]这个数组的最后一个值
而是全部里面最大的那个
HDU 2845 是这个题的升级版
http://acm.hdu.edu.cn/showproblem.php?pid=2845
*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <cmath>
#include <stack>
#include <string>
#include <sstream>
#include <map>
#include <set>
#define pi acos(-1.0)
#define LL long long
#define ULL unsigned long long
#define inf 0x3f3f3f3f
#define INF 1e18
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;
typedef pair<int, int> P;
const double eps = 1e-10;
const int maxn = 1e5 + 5;
const int N = 1e4 + 5;
const int mod = 1e8;
int a[maxn];
int cnt[maxn];
LL val[maxn], dp[maxn];
int main(void)
{
// freopen("in.txt","r", stdin);
int n, maxx;
while (cin >> n){
memset(cnt, 0, sizeof(cnt));
memset(dp, 0, sizeof(dp));
maxx = -1;
for (int i = 1; i <= n; i++){
cin >> a[i];
cnt[a[i]]++;
maxx = max(maxx, a[i]);
}
for (int i = 1; i <= maxx; i++)
val[i] = (LL)cnt[i] * (LL)i;
memset(dp, 0, sizeof(dp));
dp[1] = val[1];
for (int i = 2; i <= maxx; i++)
dp[i] = max(dp[i-1], dp[i-2]+val[i]);
LL ans = -1;
for (int i = 1; i <= maxx; i++)
ans = max(ans, dp[i]);
cout << ans << endl;
}
return 0;
}