[DP]CF455A Boredom

A. Boredom

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.

正确解法：

刚开始以为是一个区间dp，以为是把a[k]删去,a[k+1]和a[k-1]的数也要删去。

但是开的区间数组1e5开不下，去看题解，以为有更好的方法。

于是……题目看错了

是把某个数a删去，a+1和a-1也要删去。

我们设f[i]为最大值为i的最大价值。

若i不被删的话，f[i]=f[i-1]

若把i删除，f[i]=f[i-2]+cnt[i]*i； 　　

（i删除后，i-1就不能选择，就要从f[i-2]开始转移）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdlib>
#include<stack>
#define de printf("
debug:
")
#define End printf("
end

")
#define fi first
#define se second
#define P pair< int, int >
#define PII pair< pair<int, int> ,int>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
const ll mod=1000000007;
const int N=100000+100;
const int inf=0x7fffffff;
int n,a[N],maxx;
ll cnt[N];
ll f[N];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        maxx=max(maxx,a[i]);
        cnt[a[i]]++;
    }
    f[0]=0;
    f[1]=cnt[1];
    for(int i=2;i<=maxx;i++)
        f[i]=max(f[i-1],f[i-2]+cnt[i]*i);
    printf("%lld
",f[maxx]);



    return 0;
}