A. 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.
分析：
题意： 给定一个序列，每次从序列中选出一个数ak,获得ak的得分， 同时删除序列中所有的ak1,ak+1, 求最大得分的值。
思路： 存下每个数的个数放在c中，消除一个数i,会获得c[i]*i的值（因为可以消除c[i]次）， 如果从0的位置开始向右消去，那么，消除数i时，i-1可能选择了消除，也可能没有， 如果消除了i-1，那么i值就已经不存在，dp[i] = dp[i-1]， 如果没有被消除，，那么dp[i] = dp[i-2]+ c[i]*i。
LL;const double pi = acos(-1.0);mod = 1e9 + 7;;LL dp[100005];LL c[100005];int main(){memset(dp,0,sizeof(dp));memset(c,0,sizeof(c));int N;scanf(“%d”,&N);LL t;LL MAX = -1;for(int i = 1;i <= N;i++){scanf(“%I64d”,&t);MAX = max(MAX,t);c[t]++;}dp[0] = 0;dp[1] = a[1];for(int i = 2;i <= MAX;i++)dp[i] = max(dp[i – 1],dp[i – 2] + i * c[i]);printf(“%I64d\n”,dp[MAX]);return 0;}
与一个赏心悦目的人错肩，真真实实的活着，也就够了。
