cf455A boredom
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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 + 1and 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
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.
思路:dp。就是表示每一步dp的状态,即此步状态可能是前一步dp[i-1]的状态(即这一步的数已被消除),或者是前两步dp[i-2]的状态加上这一步的值(因为下一步的状态不确定,所以下一步的值不能消去)。
所以状态转移方程就是dp[i] = max(dp[i-2] + a[i]*i,dp[i-1]);
#include<bits/stdc++.h> using namespace std; const int N = 1e5+10; typedef long long ll; ll a[N],dp[N]; int main() { int n,x,ma = 0,mi = N; memset(a,0,sizeof(a)); memset(dp,0,sizeof(dp)); cin>>n; for(ll i = 0; i < n; i++) { scanf("%d",&x); a[x]++; if(x > ma) { ma = x; } if(x < mi) { mi = x; } } dp[mi] = a[mi]*mi; for(ll i = mi+1; i <= ma; i++) { dp[i] = max(dp[i-2]+a[i]*i,dp[i-1]); } cout<<dp[ma]<<endl; return 0; }
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