标签：mes for game second contain std section get lex
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 }
标签：mes for game second contain std section get lex
原文地址：http://www.cnblogs.com/zgglj-com/p/7745682.html
