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.
Dp,设数i出现a[i]次
非常easy发现,从n取到i(i全取)的最优值f{i}仅仅与f(i+1)和f(i+2)相关
#include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<functional> #include<iostream> #include<cmath> #include<cctype> #include<ctime> using namespace std; #define For(i,n) for(int i=1;i<=n;i++) #define Fork(i,k,n) for(int i=k;i<=n;i++) #define Rep(i,n) for(int i=0;i<n;i++) #define ForD(i,n) for(int i=n;i;i--) #define RepD(i,n) for(int i=n;i>=0;i--) #define Forp(x) for(int p=pre[x];p;p=next[p]) #define Lson (x<<1) #define Rson ((x<<1)+1) #define MEM(a) memset(a,0,sizeof(a)); #define MEMI(a) memset(a,127,sizeof(a)); #define MEMi(a) memset(a,128,sizeof(a)); #define INF (2139062143) #define F (100000007) #define MAXN (100000+10) long long mul(long long a,long long b){return (a*b)%F;} long long add(long long a,long long b){return (a+b)%F;} long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;} typedef long long ll; int n,a[MAXN]={0}; ll f[MAXN]={0}; int main() { // freopen("seq.in","r",stdin); // freopen(".out","w",stdout); MEM(a) cin>>n; int p,m=0; For(i,n) { scanf("%d",&p); a[p]++; m=max(m,p); } ll ans=0; ForD(i,m) { f[i]=max(f[i+1],f[i+2]+(ll)i*a[i]); ans=max(ans,f[i]); } cout<<ans<<endl; return 0; }
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