1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 1e5+10;
 5 inline ll read(){
 6     ll x=0,f=1;char ch=getchar();
 7     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 8     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 9     return x*f;
10 }
11
12 ll a[maxn],dp[maxn];
13
14 int main(){
15     int n;
16     cin >> n;
17     for(int i=1; i<=n; i++){
18         ll x; cin >> x;
19         a[x]++;
20     }
21
22     memset(dp,0,sizeof(dp));
23
24     for(int i=1; i<maxn; i++){
25         dp[i] = max(dp[i-1],dp[i-2]+a[i]*i);
26     }
27
28     cout << dp[maxn-1] << endl;
29
30     return 0;
31 }

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 1e5+10;
 5 const int INF = 0x3f3f3f3f;
 6 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 7 inline ll read(){
 8     ll x=0,f=1;char ch=getchar();
 9     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
10     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
11     return x*f;
12 }
13 //
14
15 ll a[maxn],dp[maxn][2];
16 int mx=-1;
17
18 int main(){
19     int n; n = read();
20     for(int i=1; i<=n; i++){
21         int x; x = read();
22         a[x]++;
23         mx = max(x,mx);
24     }
25
26     for(int i=1; i<=mx; i++){
27         dp[i][1] = dp[i-1][0]+a[i]*i;    // dp[i][1] 表示这个数删
28         dp[i][0] = max(dp[i-1][1],dp[i-1][0]); // dp[i][0] 表示这个数不删
29     }
30
31     cout << max(dp[mx][0],dp[mx][1]) << endl;
32     return 0;
33 }

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 1e5+10;
 5 const int INF = 0x3f3f3f3f;
 6 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 7 inline ll read(){
 8     ll x=0,f=1;char ch=getchar();
 9     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
10     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
11     return x*f;
12 }
13 //
14
15 ll a[maxn],dp[2][2];
16 int mx=-1;
17
18 int main(){
19     int n; n = read();
20     for(int i=1; i<=n; i++){
21         int x; x = read();
22         a[x]++;
23         mx = max(x,mx);
24     }
25
26     int now = 0, pre = 1;;
27     for(int i=1; i<=mx; i++){
28
29         dp[now][1] = dp[pre][0]+a[i]*i;    // dp[now][1] 表示这个数删
30         dp[now][0] = max(dp[pre][1],dp[pre][0]); // dp[now][0] 表示这个数不删
31         swap(now, pre);
32     }
33
34     dp[now][0] = max(dp[now][0],dp[pre][0]);
35     dp[now][1] = max(dp[now][1],dp[pre][1]);
36
37     ll ans = max(dp[now][0],dp[now][1]);
38     cout << ans << endl;
39
40     return 0;
41 }