求最大的连续不超过k的子序列的和。

用单调队列维护。

先求出s[1..i]的和，将前k个添加到n的结尾就相当于有循环和了。

那么对于某个sj,他的最大的序列和为s[j] - s[i]，其中  j - k - 1 <= i <= j - 1.

那么用单调队列去维护i，可以在O(1)的时间去求出s[i]。

今后有任何优化问题需要减去前面最小或者加上最大和的都可以使用单调队列去维护。

AC代码：

#include <cstdio>
#include <cstring>const int MAX_NUMBER = 1000006;
const int INF = 2000000007;int sums[2 * MAX_NUMBER];
int value[MAX_NUMBER];
int queue[2 * MAX_NUMBER];int main() {int test_case;scanf("%d", &test_case);while (test_case--) {int n, length, head, tail, max_number, max_length, start, end;scanf("%d%d", &n, &length);sums[0] = 0;head = tail = 1;queue[head] = 0;max_number = -INF;start = end = MAX_NUMBER;max_length = MAX_NUMBER;for (int i = 1; i <= n; i++) {scanf("%d", &value[i]);sums[i] = sums[i - 1] + value[i];}for (int i = n + 1; i <= n + length - 1; i++) {sums[i] = sums[i - 1] + value[i % n];}for (int i = 1; i <= n + length - 1; i++) {while (tail >= head && i - queue[head] > length) {head++;}int temp = sums[i] - sums[queue[head]];if (temp >= max_number) {if (temp > max_number) {max_number = temp;max_length = i - queue[head];start = queue[head] + 1;end = i;}else {if (queue[head] + 1 < start) {max_length = i - queue[head];start = queue[head] + 1;end = i;}else if (queue[head] + 1 == start) {if (max_length > i - queue[head]) {max_length = i - queue[head];start = queue[head] + 1;end = i;}}}}while (tail >= head && sums[queue[tail]] > sums[i]) {tail--;}queue[++tail] = i;}if (end > n) {end -= n;}printf("%d %d %d\n", max_number, start, end);}return 0;
}