A. Boredom
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.
正确解法：
刚开始以为是一个区间dp，以为是把a[k]删去,a[k+1]和a[k-1]的数也要删去。
但是开的区间数组1e5开不下，去看题解，以为有更好的方法。
于是……题目看错了
是把某个数a删去，a+1和a-1也要删去。
我们设f[i]为最大值为i的最大价值。
若i不被删的话，f[i]=f[i-1]
若把i删除，f[i]=f[i-2]+cnt[i]*i；
（i删除后，i-1就不能选择，就要从f[i-2]开始转移）
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #include<string> 6 #include<map> 7 #include<set> 8 #include<vector> 9 #include<queue> 10 #include<cmath> 11 #include<cstdlib> 12 #include<stack> 13 #define de printf("\ndebug:\n") 14 #define End printf("\nend\n\n") 15 #define fi first 16 #define se second 17 #define P pair< int, int > 18 #define PII pair< pair<int, int> ,int> 19 #define INF 0x3f3f3f3f 20 using namespace std; 21 typedef long long ll; 22 const ll mod=1000000007; 23 const int N=100000+100; 24 const int inf=0x7fffffff; 25 int n,a[N],maxx; 26 ll cnt[N]; 27 ll f[N]; 28 int main() 29 { 30 scanf("%d",&n); 31 for(int i=1;i<=n;i++) 32 { 33 scanf("%d",&a[i]); 34 maxx=max(maxx,a[i]); 35 cnt[a[i]]++; 36 } 37 f[0]=0; 38 f[1]=cnt[1]; 39 for(int i=2;i<=maxx;i++) 40 f[i]=max(f[i-1],f[i-2]+cnt[i]*i); 41 printf("%lld\n",f[maxx]); 42 43 44 45 return 0; 46 }