codeforces Round #260(div2) C解题报告

C. Boredom

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Sample test(s)

input

2
1 2

output

2

input

3
1 2 3

output

4

input

9
1 2 1 3 2 2 2 2 3

output

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.

题目大意：

给出N个数字，范围在1~10^5之内，现在做一个游戏，每次抽出一个数字a，将a删掉并计入分数，同时删掉所有等于a-1和a+1的数字。现在求分数最大。

解法：

很简单的一道DP，首先统计每个数字总共有多少个。

      由于抽a会将a-1和a+1都删掉，为了避免后效性，我们将统一方向，如若选择a，则a-1均被删除，a+1不管，因为我们是从1~maxn。我们设f[i]为选择到数字i的时候，可以获得的最大分数，这里有两种选择，一种是该层数字不选择，直接选择前一个数字，那么该层数字就会被删除即不计入分数，如若选择前2个数字，则可以选择该层数字。

状态转移方程：f[i] = max(f[i-1], f[i-2]+a[i]*i); 

代码：

#include <cstdio>
#include <algorithm>
#define N_max 123456
#define LL long long

using namespace std;

int n;
LL a[N_max], f[N_max], ans, tmp;

void init() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%I64d", &tmp);
a[tmp]++;
}
}

void solve() {
f[1] = a[1];
ans = f[1];

for (int i = 2; i < N_max; i++) {
f[i] = max(f[i-1], f[i-2]+a[i]*i);
ans = max(f[i], ans);
}

printf("%I64d\n", ans);
}

int main() {
init();
solve();
}