Codeforces 456C - Boredom(簡單DP)
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 a
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
InputThe first line contains integer n
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105).
OutputPrint a single integer — the maximum number of points that Alex can earn.
Examples input2output1 2
2input
3output
1 2 3
4input
9output
1 2 1 3 2 2 2 2 3
10Note
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.
題意:
每次從n個數中選中數值為Ak將其刪去,並將值為Ak-1的數和值為Ak+1的數全部刪去,得到Ak的分數。問最大可獲得的分數。
題解:
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; }
Codeforces 456C - Boredom(簡單DP)