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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
2 output
2 input
3 output
4 input
9 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. 题意: 每次从n个数中选中数值为Ak将其删去,并将值为Ak-1的数和值为Ak+1的数全部删去,得到Ak的分数。问最大可获得的分数。 题解: DP一下就好了。
#include<iostream> #include<cstring> #include<algorithm> using namespace std; const int maxn=1e5+5; long long cnt[maxn],dp[maxn]; int main() { int n; while(cin>>n) { memset(cnt,0,sizeof(cnt)); for(int i=0;i<n;i++) { int x; cin>>x; cnt[x]++; } dp[0]=0,dp[1]=cnt[1]; long long ans=0; for(int i=2;i<maxn;i++) { dp[i]=max(dp[i-1],dp[i-2]+cnt[i]*i); ans=max(ans,dp[i]); } cout<<ans<<endl; } return 0; }
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