CodeForces
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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.
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).
Print a single integer — the maximum number of points that Alex can earn.
21 2
2
31 2 3
4
91 2 1 3 2 2 2 2 3
10
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.
典型的dp问题
#include<stdio.h>#include<algorithm>#include<string.h>#define maxn 100010using namespace std;int main(){ long long ans; int n,i,num,Max; while(scanf("%d",&n)!=EOF) { long long a[maxn],dp[maxn]; Max=0; memset(a,0,sizeof(a)); memset(dp,0,sizeof(dp)); for(i=0;i<n;i++) { scanf("%d",&num); a[num]++; if(num>Max) Max=num; } dp[0]=0; dp[1]=a[1]; for(i=2;i<=Max;i++) { dp[i]=max(dp[i-1],dp[i-2]+i*a[i]); } printf("%lld\n",dp[Max]); }}
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