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#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

 

int n,k;

int minn = 1e9;//最小值

int ltor,rtor,pt; //左力矩、右力矩、切割點

vector<int> prex;//前綴和

vector<int> sufx;//後綴和

 

 

LL solve(vector<int> &lis,int st,int ed,int k){

//cout<<st<<" "<<ed<<endl;

if(k==0) return 0;

if(abs(ed-st) < 3) return 0;

else{

prex.clear();

prex.push_back(lis[st]);

for(int i=st+1;i<ed;i++) prex.push_back(lis[i]+prex[i-1]);

sufx.clear();

sufx.push_back(prex[ed-st-1]);

for(int i=st+1;i<ed;i++)sufx.push_back(sufx[i-st-1]-lis[i-1]);

minn = 1e9;

ltor = 0;rtor = 0;

for(int i=st+1;i<ed;i++)rtor += lis[i]*(i-st);

pt = 0;

for(int i=st+1;i<ed-1;i++){//支點範圍

ltor += prex[i-1];

rtor -= sufx[i];

cout<<"ltor = "<<ltor<<", rtor = "<<rtor<<endl;

if(abs(rtor - ltor)<minn){

minn = abs(rtor - ltor);

pt = i;

}

}

//cout<<st<<" "<<pt<<" "<<pt+1<<" "<<ed<<endl;

}

return lis[pt] + solve(lis,st,pt,k-1) + solve(lis,pt+1,ed,k-1);

}

 

int main(){

ios_base::sync_with_stdio(false);cin.tie(0);

cin>>n>>k;

vector<int> arr(n);

for(int i=0;i<n;i++) cin>>arr[i];

cout<<solve(arr,0,n,k);

}

我的想法是
以範例測資為例

7 3
2 4 1 3 7 6 9

我以為會分成
(2),4,(1,3),7,(6,9)
結果我發現會多出
(1,3,7,6,9)

我不知道如何解決