Boredom easy dp

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 455A

Description

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_kpoints 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, ..., a_n (1 ≤ a_i ≤ 10⁵).

Output

Print a single integer — the maximum number of points that Alex can earn.

Sample Input

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

Hint

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.

My code：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
bool cmp(int x,int y)
{
    return x>y;
}

int main()
{
    int n;
    long long dp[100005];
    int a[100005],b[100005],m[100005];
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        memset(m,0,sizeof(m));
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(m[a[i]]>0)
            {
                m[a[i]]++;
                a[i]=0;i--;n--;
            }
            else
            {
                m[a[i]]++;
            }
        }
        /*printf("%d
",n);
        for(i=1;i<=n;i++)
            printf("%d ",a[i]);*/

        sort(a+1,a+n+1,cmp);
        dp[0]=0;
        dp[1]=(long long)a[1]*m[a[1]];
        if(a[1]-1==a[2])
            dp[2]=(long long)a[2]*m[a[2]];
        else
            dp[2]=dp[1]+(long long)a[2]*m[a[2]];
        //printf("%I64d %I64d
",dp[1],dp[2]);
        for(i=3;i<=n;i++)
        {
            long long maa=0;
            maa=max(dp[i-2],dp[i-3]);
            if(a[i-1]-1==a[i])
                dp[i]=maa+(long long)a[i]*m[a[i]];
            else
            {
                dp[i]=(long long)a[i]*m[a[i]]+max(maa,dp[i-1]);
            }
        }
        long long ans=0;
        for(i=1;i<=n;i++)
            if(ans<dp[i])
                ans=dp[i];
        printf("%I64d
",ans);
    }
    return 0;
}

别宁家的代码 短小&精悍

#include <cstdio>  
#include <cstring>  
#include <algorithm>  
using namespace std;  
#define LL __int64  
const int MAXN = 100017;  
LL c[MAXN], dp[MAXN];  
void init()  
{  
    memset(dp,0,sizeof(dp));  
    memset(c,0,sizeof(c));  
}  
int main()  
{  
    LL n;  
    LL tt;  
    int i, j;  
    while(~scanf("%I64d",&n))  
    {  
        init();  
        int maxx = 0;  
        for(i = 1; i <= n; i++)  
        {  
            scanf("%I64d",&tt);  
            if(tt > maxx)  
                maxx = tt;  
            c[tt]++;  
        }  
        dp[0] = 0, dp[1] = c[1];  
        for(i = 2; i <= maxx; i++)  
        {  
            dp[i] = max(dp[i-1],dp[i-2]+c[i]*i);  
        }  
        printf("%I64d
",dp[maxx]);  
    }  
    return 0;  
}