Boredom CodeForces - 455A
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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.
Example
2
1 2
2
3
1 2 3
4
9
1 2 1 3 2 2 2 2 3
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.
题解:dp[ i ][ 0~1 ],表示以i结尾(0表示不选i,1表示1选i)所得到的最大值。状态定好了,转移方程就好想了
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 5 const int maxn=1e5+5; 6 7 ll n,ma; 8 ll a[maxn],dp[maxn][2]; 9 10 int main() 11 { ma=0; 12 cin>>n; 13 for(int i=1;i<=n;i++){ 14 ll temp; 15 cin>>temp; 16 a[temp]++; 17 ma=max(ma,temp); 18 } 19 for(int i=1;i<=ma;i++) a[i]=(ll)i*a[i]; 20 21 dp[0][0]=dp[0][1]=0; 22 for(int i=1;i<=ma;i++){ 23 dp[i][0]=max(dp[i-1][1],dp[i-1][0]); 24 dp[i][1]=dp[i-1][0]+a[i]; 25 } 26 cout<<max(dp[ma][0],dp[ma][1])<<endl; 27 }
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