이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
// oooo
/*
har chi delet mikhad bebar ~
gitar o ba khodet nabar! ~
;Amoo_Hasan;
*/
#include<bits/stdc++.h>
//#pragma GCC optimize("O3,no-stack-protector,unroll-loops")
//#pragma GCC target("avx2,fma")
using namespace std;
typedef long long ll;
typedef long double ld;
#define Sz(x) int((x).size())
#define All(x) (x).begin(), (x).end()
#define wtf(x) cout<<#x <<" : " <<x <<endl
#define mak make_pair
constexpr ll INF = 1e18, N = 1e5 + 10, MAXK = 2e2 + 10;
ll a[N], ps[N], dp[N][MAXK];
pair<ll, ll> par[N][MAXK];
bool mark[N];
int main() {
ios :: sync_with_stdio(0), cin.tie(0);
int n, k; cin >>n >>k;
for(int i = 1; i <= n; i++) cin >>a[i];
for(int i = 1; i <= n; i++) ps[i] = ps[i - 1] + a[i];
memset(dp, 63, sizeof(dp));
dp[0][0] = 0;
/*
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= min(k + 1, i); j++) {
ll val = INF;
for(int x = i - 1; x >= 0; x--) {
if(dp[x][j - 1] >= INF) continue;
ll cnt = dp[x][j - 1] + (ps[i] - ps[x]) * (ps[i] - ps[x]);
if(cnt < val) {
val = cnt;
par[i][j] = mak(x, j - 1);
}
}
dp[i][j] = val;
}
}
*/
/* ll ans = ps[n] * ps[n];
ans -= dp[n][k + 1];
ans /= 2;
cout<<ans <<endl;
*/
/*
vector<int> pos;
pair<ll, ll> p = mak(n, k + 1);
while(p.second > 0) {
p = par[p.first][p.second];
if(p.first > 0) pos.push_back(p.first);
}
while(!pos.empty()) {
cout<<pos.back() <<" ";
pos.pop_back();
}*/
return 0;
}
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