Codeforces Round #260 (Div. 2) C. Boredom【DP】

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

21 2

output

2

input

31 2 3

output

4

input

91 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.

输入的时候 记录每个数出现的次数，遍历所有数，对每个数有选和不选的情况，然后...

一维数组：dp[i]表示 i 这个数处理的结果，如果选，就是 dp[ i - 2 ] +这个数的值*数量，因为如果 i - 1 被选了，i 就自然被删除了，如果不选，就是 dp [ i - 1 ] 

#include<stdio.h>#include<iostream>#include<algorithm>using namespace std;#define MAX 111111#define ll __int64ll sum[MAX],dp[MAX];int main(){    int n,x,nmax = 0;    scanf("%d",&n);    for(int i=0;i<n;i++)    {        scanf("%d",&x);        sum[x] += x;        nmax = max(nmax,x);    }    dp[0] = 0;    dp[1] = sum[1];    for(int i=2;i<=nmax;i++)    {        dp[i] = max(dp[i-2]+sum[i],dp[i-1]);    }    printf("%I64d\n",dp[nmax]);}

二维：不选的话就是上一个数选或不选的最大值，选的话是上一个数不选 + 这个数的值*数量

#include<stdio.h>#include<iostream>#include<algorithm>using namespace std;#define MAX 111111#define ll __int64ll sum[MAX],dp[MAX][2];int main(){    int n,x,nmax = 0;    scanf("%d",&n);    for(int i=0;i<n;i++)    {        scanf("%d",&x);        sum[x]++;        nmax = max(nmax,x);    }    for(int i=1;i<=nmax;i++)    {        dp[i][0] = max(dp[i-1][0],dp[i-1][1]);        dp[i][1] = dp[i-1][0] + sum[i]*i;    }    printf("%I64d\n",max(dp[nmax][0],dp[nmax][1]));}

上面的代码都用全局变量方式初始化了0，所以多次输入的话要memset