Coderforces-455A
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.
InputThe 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.
2 1 2
2
3 1 2 3
4
9 1 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.
题意:就是给你N个数。你可选择大小的为A的数,但是A-1和A+1必须从数队移走。
题解:本题与数的顺序无关(这时候想到了DP),我们先对其排序。然后找其状态转移方程。假设dp[j]表示从1加到j 时最大值。则dp[j]=max(dp[j-1],dp[j-2]+i*cnt[i]).【ps:cnt[i]表示数值为】
AC代码为:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
long long n, a[100002], cot[100002], d[100002];
int main()
{
cin >> n;
long long Max = 0;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
if (a[i] > Max)
Max = a[i];
cot[a[i]]++;
}
d[1] = cot[1]; d[0] = 0;
for (int i = 2; i <= Max; i++)
d[i] = max(d[i - 1], d[i - 2] + cot[i] * i);
cout << d[Max] << endl;
return 0;
}