最大连续和（单调队列）

区间连续和——考虑前缀和

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int n,m;
int s[300090];
int q[300090];
int main() {
	int x;
	int ans=-0xfffffff;
	scanf ("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf ("%d", &x);
		s[i] = s[i - 1] + x;
	}
	for (int i = 1; i <= n; i++) 
		for (int j = i - m; j < i; j++) 
			ans = max (ans, s[i]-s[j]);
	printf ("%d", ans);
	return 0;
	
}

两层循环非吾愿 单调队列化二层循环

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int n,m;
int s[300090];
int q[300090];
int num[300090];
int main() {
	int x;
	int head = 1, tail = 1; 
	int ans = -0xfffffff;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &x);
		s[i] = s[i - 1] + x;
	}
	for (int i = 1; i <= n; i++) {
		while(head<=tail&&num[head]<i-m) head++;
		ans = max (ans, s[i] - q[head]);
		while(q[tail] >= s[i]&&head <= tail) tail--;
		q[++tail] = s[i];
		num[tail] = i;
	} 	
	printf("%d", ans);
	return 0;
	
}

此题与上题相似

上题m个里取1个 队列存前m个的最小值 +w[i]

此题连续取m个  队列存前m个的前缀和最小值 w[i]-..