題目連結：
HDU 3415 Max Sum of Max-K-sub-sequence
題意：
給n個數，首尾相連，求長度不超過k的最大連續子序列和。
資料範圍：1≤n≤100000,1≤k≤n
分析：
因為考慮首尾相連，所以我們把n個數看成2∗n個數，其中data[n+]=data[i]。求個字首和sum[]。我們考慮以i為結尾的長度不超過k的最大連續子序列和。顯然是sum[i]−min(sum[j]),j∈[i−k,i−1].那麼問題就轉化為求區間最小值了，所以我們維護一個單調非遞增佇列即可。注意下邊界。

#include <stdio.h>
#include <string.h>
 

#include <algorithm>
#include <math.h>
#include <climits>
using namespace std;
typedef long long ll;
const int MAX_N = 100010 * 2;

int T, n, k, head, tail, st, ed;
ll data[MAX_N], sum[MAX_N]; 
int que[MAX_N];

int main()
{
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d"
 
, &n, &k);
        sum[0] = 0;
        for (int i = 1; i <= n; ++i) {
            scanf("%lld", &data[i]);
            data[i + n] = data[i];
        }
        ll ans = data[1], tmp;
        st = 1, ed = 1;
        for (int i = 1; i <= 2 * n; ++i) {
            sum[i] = sum[i - 1] + data[i];
            if
 
 (i <= k) {
                if (sum[i] > ans) {
                    ans = sum[i], ed = i;
                }
            }
        }
        // 維護單調遞增佇列，存字首和
        head = tail = 0;
        que[tail++] = 1;
        for (int i = 2; i <= n + k; ++i) {
            // 剔除在i之前距離超過k的字首和
            while (head != tail && i - que[head] > k) {
                ++head;
            }
            // 剔除在i之前字首和大的字首和
            while (head != tail && sum[i] <= sum[que[tail - 1]]) {
                --tail;
            }
            int flag = -1;
            if (i <= k && sum[que[head]] >= 0) {
                tmp = sum[i], flag = 1;
            } else if (head == tail) {
                tmp = data[i], flag = 2;
            } else {
                tmp = sum[i] - sum[que[head]];
                flag = 3;
            }
            if (tmp > ans) {
                ans = tmp;
                if (flag == 1) st = 1;
                else if (flag == 2) st = i;
                else st = que[head] + 1;
                ed = i;
            }
            que[tail++] = i;
            //printf("i = %d ans = %d:\n", i, ans);
            /*
            for (int j = head; j < tail; ++j) {
                printf("%d ", sum[que[j]]);
            }
            printf("\n");
            */
        }
        if (st > n) st -= n;
        if (ed > n) ed -= n;
        printf("%lld %d %d\n", ans, st, ed);
    }
    return 0;
}