The meaning of problems
You're playing on a length \ (n \) game non-negative integer sequence. This game you need to sequence into \ (k + 1 \) non-empty blocks. In order to obtain \ (k + 1 \) block, you need to repeat the following operations \ (K \) views:
- Selecting a block of more than one element (initially you only one, i.e. the entire sequence)
- Select two elements adjacent to this block is parted in the middle to give two non-empty blocks.
After each operation you will get two points newly generated element blocks and of the product. You want to maximize the final total score.
Solution
Obviously each pair \ (i, J \) , if they are not in the same block, will produce \ (a_i * a_j \) contribution.
Apparently \ (\ Large f_ {p, i} = \ max ({f_ {p-1, j}} + s_j (s_i-s_j)) \)
Provided transfer \ (J, k \) , than k J
\(j<k\)
\(\large f_{p-1,j}+(s_j*(s_i-s_j))>f_{p-1,k}+(s_k*(s_i-s_k))\)
\(\large f_{p-1,j}-s_j^2+s_is_j>f_{p-1,k}-s_k^2+s_is_k\)
\(\huge s_i>\frac{f_{p-1,k}- f_{p-1,j}+s_j^2-s_k^2}{s_j-s_k}\)
Directly on the slope can be optimized
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define IL inline
#define RG register
#define gi geti<int>()
#define gl geti<ll>()
#define gc getchar()
#define File(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
template<typename T>IL bool chkmax(T &x,const T &y){return x<y?x=y,1:0;}
template<typename T>IL bool chkmin(T &x,const T &y){return x>y?x=y,1:0;}
template<typename T>
IL T geti()
{
RG T xi=0;
RG char ch=gc;
bool f=0;
while(!isdigit(ch))ch=='-'?f=1:f,ch=gc;
while(isdigit(ch))xi=xi*10+ch-48,ch=gc;
return f?-xi:xi;
}
template<typename T>
IL void pi(T k,char ch=0)
{
if(k<0)k=-k,putchar('-');
if(k>=10)pi(k/10);
putchar(k%10+'0');
if(ch)putchar(ch);
}
const int N=1e5+7;
int n,k;
ll a[N],f[N][2],s[N];
int pr[N][210],now,pre;
int q[N];
inline ll sqr(ll p){return p*p;}
inline double slope(int j,int k)
{
if(s[j]==s[k])return -1e18;
return (f[k][pre]-f[j][pre]+sqr(s[j])-sqr(s[k]))/static_cast<double>(s[j]-s[k]);
}
int main(void)
{
n=gi,k=gi;
for(int i=1;i<=n;++i)a[i]=gi,s[i]=s[i-1]+a[i];
for(int p=1;p<=k;++p)
{
q[0]=0;
int l=0,r=0;
now=p&1,pre=now^1;
for(int i=1;i<=n;++i)
{
while(l<r&&slope(q[l],q[l+1])<=s[i])++l;
int j=q[l];
f[i][now]=f[j][pre]+(s[i]-s[j])*s[j];
pr[i][p]=j;
while(l<r&&slope(q[r],i)<=slope(q[r-1],q[r]))--r;
q[++r]=i;
}
}
pi(f[n][now],'\n');
for(int i=k,w=n;i;--i)pi(w=pr[w][i],' ');
return 0;
}