이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
// In The Name Of The Queen
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100005, K = 209;
int n, k, lvl, P[K][N];
ll A[N], dp[2][N];
inline ll Get(int l, int r)
{
return ((A[r] - A[l]) * (A[r] - A[l]));
}
void Solve(int l, int r, int le, int ri, int j)
{
if (r < l) return ;
int md = l + r >> 1, opt = -1;
dp[j][md] = (ll)2e18;
for (int i = le; i <= min(ri, md); i ++)
if (dp[j][md] >= dp[!j][i - 1] + Get(i - 1, md))
dp[j][md] = dp[!j][i - 1] + Get(i - 1, md), opt = i;
P[lvl][md] = opt - 1;
Solve(l, md - 1, le, opt, j);
Solve(md + 1, r, opt, ri, j);
}
int main()
{
scanf("%d%d", &n, &k); k ++;
for (int i = 1; i <= n; i ++)
scanf("%lld", &A[i]), A[i] += A[i - 1];
memset(dp, 63, sizeof(dp));
dp[0][0] = 0;
for (lvl = 1; lvl <= k; lvl ++)
Solve(1, n, 1, n, lvl & 1), dp[0][0] = (ll)2e18;
printf("%lld\n", (A[n] * A[n] - dp[k & 1][n]) / 2);
vector < int > V;
for (; k > 1; k --)
V.push_back(P[k][n]), n = P[k][n];
reverse(V.begin(), V.end());
for (int i : V)
printf("%d ", i);
return 0;
}
컴파일 시 표준 에러 (stderr) 메시지
sequence.cpp: In function 'void Solve(int, int, int, int, int)':
sequence.cpp:15:16: warning: suggest parentheses around '+' inside '>>' [-Wparentheses]
int md = l + r >> 1, opt = -1;
~~^~~
sequence.cpp: In function 'int main()':
sequence.cpp:26:10: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
scanf("%d%d", &n, &k); k ++;
~~~~~^~~~~~~~~~~~~~~~
sequence.cpp:28:29: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
scanf("%lld", &A[i]), A[i] += A[i - 1];
~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~
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