codeforce 455A. Boredom（经典dp水题）
A. 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.
Examples
input
Copy
2
1 2
output
Copy
2
input
Copy
3
1 2 3
output
Copy
4
input
Copy
9
1 2 1 3 2 2 2 2 3
output
Copy
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
对于当前数字的取舍，取决于前一个数字的和与当前+前前个数字的和 的大小
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int MAXN = 2e6+5;
int n,m,flag,flag1,ma,mi,ans,sum,T,k;
ll a[MAXN],b[MAXN],dp[MAXN];
char c[MAXN];
int mapp[5001][5001];
string str,str1,str2;
set ss;
vector v;
bool judge(int x){if(x==2||x==3)return false;for(int i=2;i<=sqrt(x);i++){if(x%i==0)return true;}return false;}
struct node
{
int a,b;
}s[1001];
int main()
{
cin>>n;
memset(a,0,sizeof(a));
memset(dp,0,sizeof(dp));
int ma=0;
for(int i=1;i<=n;i++)
{
int t;
cin>>t;
a[t]+=t;
if(t>ma)
ma=t;
}
dp[1]=a[1];
for(int i=2;i<=ma;i++)
{
dp[i]=max(dp[i-1],dp[i-2]+a[i]);
}
cout<