이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <assert.h>
#include <algorithm>
#include <vector>
#include <set>
#include <string>
#include <queue>
#include <map>
#include <bits/stdc++.h>
#define initrand mt19937 mt_rand(time(0));
#define rand mt_rand()
#define int long long
#define INF 10000000000000000
using namespace std;
#define mid(l, u) ((l+u)/2)
//#define cin fin
//#define cout fout
const int N = 100005;
const int K = 205;
int pref[N];
int n, k;
int dp[2][N];
signed nxt[K][N];
stack<pair<pair<int, int>, pair<int, int>>> toEval;
void f(int currk, int l, int r, int ll, int rr){ //dp(currk, m) = best cost when you want to make currk splits with first split starting at m
if(l>r){
return;
}
int m = mid(l, r);
//we want to calculate dp[currk][m]
//look at dp[currk-1][max(ll, m+1)] to dp[currk - 1][min(rr, n - currk)].
//The answer for dp[curr-1][x] will be (sum(m to x-1) * sum(x to n)) + dp[curr-1][x]
//base case - dp[0][anything] = 0;
//cout<<currk<<" "<<m<<" "<<ll<<" "<<rr<<endl;
//n - x + 1 >= currk
for(int x = max(ll, m+1);x<=min(rr, n - (currk - 1));x++){
int tempans = ((pref[x-1] - pref[m-1]) * (pref[n] - pref[x-1])) + dp[0][x];
//cout<<"ANS AT "<<x<<" is "<<tempans<<endl;
if(tempans >= dp[1][m]){
nxt[currk][m] = x;
dp[1][m] = tempans;
}
}
//cout<<"dp["<<currk<<"]["<<m<<"] = "<<dp[currk][m]<<endl;
f(currk, l, m-1, ll, nxt[currk][m]);
f(currk, m+1, r, nxt[currk][m], rr);
}
signed main(){
//ofstream fin("2.in");
//ofstream fout("2.out");
cin>>n>>k;
for(int i = 1;i<=n;i++){
cin>>pref[i];
pref[i] += pref[i-1];
}
for(int currk = 1;currk<=k;currk++){
f(currk, 1, n - currk, 1, n);
for(int i = 1;i<=n;i++){
dp[0][i] = dp[1][i];
dp[1][i] = 0;
}
}
cout<<dp[0][1]<<endl;
int lol = k;
int currst = 1;
while(lol){
cout<<(nxt[lol][currst] - 1)<<" ";
currst = nxt[lol][currst];
lol--;
}
cout<<endl;
}
/*
7 3
4 1 3 4 0 2 3
*/
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