6 4
1 -3 5 1 -2 3

7

本题我们要找一个不超过M长度的和最大的子段(可以不取)。那么直接把所有前缀和求出来，For一遍每一个右端点，用单调队列维护一下前面m个以内最小的前缀和是多少就行了

Code:

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 1e6 + 5;
int q[maxn][2], f = 1, b = 0, n, m, sum[maxn], mx = 0;

void push(int i, int x) {
	while(q[b][1] >= x && f <= b) --b;
	q[++b][0] = i, q[b][1] = x;
	if(q[f][0] <= i - m) ++f;
}

int main() {
	scanf("%d%d", &n, &m);
	for(register int i = 1; i <= n; ++i)
		scanf("%d", &sum[i]);
	for(register int i = 2; i <= n; ++i)
		sum[i] += sum[i - 1];
	for(register int i = 1; i <= n; ++i) 
		mx = max(mx, sum[i] - q[f][1]), push(i, sum[i]);
	printf("%d", mx);
	return 0;
}