题目:
解析:
动态规划,设\(f[i]\)表示到第\(i\)位的最大值,我们枚举i之前的j个位置\((j<k)\),记录一下这\(j+1\)个数(包括自己)的最大值\(mx\),转移方程就是\(f[i]=max(f[i],f[i-j-1]+mx\times (j+1))\)
代码:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n, m, num;
int a[N], f[N];
template<class T>void read(T &x) {
x = 0; int f = 0; char ch = getchar();
while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
x = f ? -x : x;
return;
}
int main() {
read(n), read(m);
for (int i = 1; i <= n; ++i) read(a[i]);
for (int i = 1; i <= n; ++i) {
int mx = -1;
f[i] = f[i - 1] + a[i];
for (int j = 0; j < m && i - j > 0; ++j) {
mx = max(mx, a[i - j]);
f[i] = max(f[i], f[i - j - 1] + mx * (j + 1));
}
}
cout << f[n];
}