A. Boredom
题解：简单DP，用a数组保存每个数字各自出现了几次，然后对于ans[i]如果选择了这个数那么之前的只能选择ans[i-2]+i*a[i]，或者之前的ans[i-1]
#include <iostream>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+10;
const int inf = 0x7f7f7f7f;
ll a[N];
ll ans[N];
int main(){
int n;cin >> n;
int ans1 = 0,ans2 = 0;
for(int i = 1;i <= n;i++){
int x;cin >> x;
a[x]++;
}
ans[1] = a[1];
for(int i = 2;i <= 100002;i++){
ans[i] = max(ans[i-1],ans[i-2]+i*a[i]);
// cout<<ans[i]<<endl;
}
cout<<ans[100002]<<endl;
return 0;
}
