CF940E Cashback solution

This question seems to be very annoying. At first glance, I really didn't have any ideas.

But after careful analysis of the topic, we will find:

• $c=At\; 1$ , the answer is 0, but it seems that there is no such point?
• $c>When\; n$ , the answer is the sum of the sequence.
• $c=When\; n$ , the answer is the sum of the sequence minus the minimum value.
• $1<c<n$ 且 $n<2\times At\; c$ , the sequence can only be divided into one segment at this time, and other sequences can be regarded as intervals of length 1.
• $1<c<n$ 且 $2\times c\le At\; n$ , we can divide the sequence into multiple segments, while other sequences can still be divided into intervals of length 1.

So there are only two intervals: length 1 and length $c$。

Next, we mainly consider $1<c<n$ 's situation.

We found that at this time the problem has become a dp problem.

Setting $f_i$Means to $a_i$ The answer at the deadline, then:

• $1\le i<c$ 时：$$f_i=\sum _{j=1}^i{a}_j$$
• $c\le i\le n$ 时：$$f_i=\min (f_{i-c}+su{m}_i-su{m}_{i-c}-min{n}_i,f_{i-1}+a_i)$$

Explanation: $sum$ is the prefix and the array,$min{n}_i$Display $[i-c+1,i]$ The minimum value in this interval.

Wait: the minimum value of a fixed interval? Isn't this what monotonous queues do!

So monotonic queue $O\left(N\right)$ Resolution$min{n}_i$， After that $O\left(n\right)$ dp is fine.

Total complexity $O\left(n\right)$。

Code:

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int MAXN = 1e5 + 10;
int n, c, q[MAXN], l = 1, r = 0;
LL a[MAXN], f[MAXN], minn[MAXN], sum[MAXN];

LL read()
{
    
    
	LL sum = 0, fh = 1; char ch = getchar();
	while (ch < '0' || ch > '9') {
    
    if (ch == '-') fh = -1; ch = getchar();}
	while (ch >= '0' && ch <= '9') {
    
    sum = (sum << 3) + (sum << 1) + (ch ^ 48); ch = getchar();}
	return sum * fh;
}

int main()
{
    
    
	n = read(), c = read();
	for (int i = 1; i <= n; ++i) a[i] = read();
	for (int i = 1; i <= n; ++i)
	{
    
    
		while (l <= r && q[l] + c <= i) l++;
		while (l <= r && a[i] <= a[q[r]]) r--;
		q[++r] = i; minn[i] = a[q[l]]; sum[i] = sum[i - 1] + a[i];
	}
	for (int i = 1; i < c; ++i) f[i] = f[i - 1] + a[i];
	for (int i = c; i <= n; ++i) f[i] = min(f[i - c] + sum[i] - sum[i - c] - minn[i], f[i - 1] + a[i]);
	printf("%lld\n", f[n]);
	return 0;
}