LoveYuigahamaYui @ 2024-07-06 11:23:42
#include <iostream>
#include <deque>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
ll n, k, s[N], dp[N];
deque <int> q;
int main()
{
cin >> n >> k;
for (int i = 1; i <= n; ++i)
{
cin >> s[i];
s[i] += s[i-1];
}
for (int i = 0; i <= n; ++i)
{
if (q.size() && i - q.front() > k) q.pop_front();
if (i <= k) dp[i] = s[i];
else if (q.size()) dp[i] = s[i] + dp[q.front() - 1] - s[q.front()];
while (q.size() && dp[q.back() - 1] - s[q.back()] <= dp[i - 1] - s[i]) q.pop_back();
q.push_back(i);
}
cout << dp[n];
}
by _Liyx_ @ 2024-07-06 17:34:30
@LoveYuigahamaYui 在第26行后加一个dp[i]=max(dp[i-1],dp[i])
by _Liyx_ @ 2024-07-06 17:34:54
这是经验
by LoveYuigahamaYui @ 2024-07-07 15:53:53
@Liyx 多谢多谢