CodeForces - 940E Cashback

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You are given an array a of length n and an integer c.

The value of some array b of length k is the sum of its elements except for the smallest. For example, the value of the array [3, 1, 6, 5, 2] with c = 2 is 3 + 6 + 5 = 14.

Among all possible partitions of a into contiguous subarrays output the smallest possible sum of the values of these subarrays.

Input

The first line contains integers n and c (1 ≤ n, c ≤ 100 000).

The second line contains n integers a_i (1 ≤ a_i ≤ 10⁹) — elements of a.

Output

Output a single integer  — the smallest possible sum of values of these subarrays of some partition of a.

Example

Input

3 5
1 2 3

Output

6

Input

12 10
1 1 10 10 10 10 10 10 9 10 10 10

Output

92

Input

7 2
2 3 6 4 5 7 1

Output

17

Input

8 4
1 3 4 5 5 3 4 1

Output

23

Note

In the first example any partition yields 6 as the sum.

In the second example one of the optimal partitions is [1, 1], [10, 10, 10, 10, 10, 10, 9, 10, 10, 10] with the values 2 and 90 respectively.

In the third example one of the optimal partitions is [2, 3], [6, 4, 5, 7], [1] with the values 3, 13 and 1 respectively.

In the fourth example one of the optimal partitions is [1], [3, 4, 5, 5, 3, 4], [1] with the values 1, 21 and 1 respectively.

 

做法：

一道dp

设dp[i]为前i个的最小答案

那么

dp[i] = min(dp[i-1] , dp[i-c] + sum[（a[ i-c+1] ), a[i] ] - min(a[i-c+1] , a[i] ) )

 

代码:

 1 #include<iostream>
 2 using namespace std;
 3 #include<cstdio>
 4 #include<cstring>
 5 #define read(x) scanf("%I64d",&x)
 6 template<class T>
 7 class RMQ{
 8     public:
 9      T* a;
10      int t;
11      T **mn;
12      T maxn;
13      int *log2;
14      void SetMaxn(T *maxn){
15         this->maxn=*maxn;
16      }
17      void SetMaxn(T maxn){
18         this->maxn=maxn;
19      }
20      void Creat(T a[],int maxn){//建立一个最大为maxn的RMQ处理类 
21           int k=1,p=0;
22           log2=new int[maxn+10];
23           for(int i=0;i<=maxn;i++)
24               log2[i]=(i==0?-1:log2[i>>1]+1);
25           while(k<maxn){
26              k*=2;
27             p+=1;
28         }
29         t=p;
30         mn=new T*[maxn+10];
31         for(int i=0;i<maxn;++i){
32             mn[i] = new T[t+1];
33             mn[i][0]=a[i];
34             for(int j=1;j<=t;++j)
35                 mn[i][j]=maxn;
36         }
37          for(int j=1;j<=t;++j)
38              for(int i=0;i+(1<<j) <= maxn;++i){
39                      T sa=mn[i][j-1];
40                     T sb=mn[i+(1<<(j-1))][j-1];
41                      if(sa<sb)
42                          mn[i][j]=sa;
43                      else
44                          mn[i][j]=sb;
45              }    
46      }
47     T Getx(int ql,int qr){
48             int k=log2[qr-ql+1];
49             return min(mn[ql][k],mn[qr-(1<<k)+1][k]);
50     }
51     T Getw(int ql,int qr){
52             --ql,--qr;
53             int k=log2[qr-ql+1];
54             return min(mn[ql][k],mn[qr-(1<<k)+1][k]);
55     }
56 } ;
57 typedef long long LL;
58 RMQ<LL> rr;
59 LL n,c;
60 LL a[100100];
61 LL f[100100];
62 LL s[100100];
63 int main(){
64     rr.SetMaxn(1e15);
65     s[0]=0;
66     memset(f,-1LL,sizeof(f));
67     read(n);
68     read(c);
69     for(int i=1;i<=n;i++){
70         read(a[i]);
71         s[i] = s[i-1] + a[i];    
72     }
73     a[0]=1e15;
74     f[0]=0;
75     rr.Creat(a,n+1);
76     for(int i=1;i<=n;i++){
77         f[i] = f[i-1] + a[i];
78         LL zans = 1e15;
79         if(i-c>=0)
80             zans = f[i-c] + s[i] - s[i-c] - rr.Getx(i-c+1,i);
81         if (f[i]>zans)
82             f[i] = zans;
83     //    cout<<i<<" "<<f[i]<<endl;
84     }
85     printf("%I64d",f[n]);
86     return 0;
87 }

 

 

转载于:https://www.cnblogs.com/xfww/p/8468824.html