Since you are the best Wraith King, Nizhniy Magazin «Mir» at the centre of Vinnytsia is offering you a discount.
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 ai (1 ≤ ai ≤ 109) — elements of a.
Output
Output a single integer — the smallest possible sum of values of these subarrays of some partition of a.
Example
3 5
1 2 3
6
12 10
1 1 10 10 10 10 10 10 9 10 10 10
92
7 2
2 3 6 4 5 7 1
17
8 4
1 3 4 5 5 3 4 1
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 }
