操作：查询区间最大连续和。
每个节点保存当前区间往左，往右的最大和。
往左最大连续和=max（左区间往左最大连续和，左区间的和+右区间往左最大连续和）。
区间最大值=max（左，右区间最大值，左区间右连续+右区间左连续）。
返回答案时，区间需要不断的合并。
1 #include<cstdio> 2 #define MAX(a,b) ((a)>(b)?(a):(b)) 3 #define MAXN 50010 4 #define oo 1000000000 5 struct node { 6 int left, right, sum, val; 7 void Init() { 8 sum = 0; 9 left = right = val = -oo; 10 } 11 }; 12 node tree[MAXN << 2]; 13 inline void PushUp(int rt) { 14 tree[rt].left = MAX(tree[rt << 1].left, 15 tree[rt << 1].sum + tree[rt << 1 | 1].left); 16 tree[rt].right = MAX(tree[rt << 1 | 1].right, 17 tree[rt << 1 | 1].sum + tree[rt << 1].right); 18 tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum; 19 tree[rt].val = MAX(tree[rt << 1].val, tree[rt << 1 | 1].val); 20 tree[rt].val = MAX(tree[rt].val, 21 tree[rt << 1].right + tree[rt << 1 | 1].left); 22 } 23 void Build(int L, int R, int rt) { 24 if (L == R) { 25 scanf("%d", &tree[rt].val); 26 tree[rt].left = tree[rt].right = tree[rt].sum = tree[rt].val; 27 } else { 28 int mid = (L + R) >> 1; 29 Build(L, mid, rt << 1); 30 Build(mid + 1, R, rt << 1 | 1); 31 PushUp(rt); 32 } 33 } 34 node Query(int x, int y, int L, int R, int rt) { 35 if (x <= L && R <= y) 36 return tree[rt]; 37 int mid = (L + R) >> 1; 38 node a, b, res; 39 a.Init(), b.Init(), res.Init(); 40 if (x <= mid) 41 a = Query(x, y, L, mid, rt << 1); 42 if (y > mid) 43 b = Query(x, y, mid + 1, R, rt << 1 | 1); 44 res.left = MAX(a.left, a.sum + b.left); 45 res.right = MAX(b.right, b.sum + a.right); 46 res.sum = a.sum + b.sum; 47 res.val = MAX(a.val, b.val); 48 res.val = MAX(res.val, a.right + b.left); 49 return res; 50 } 51 int main() { 52 int n, q, x, y; 53 while (~scanf("%d", &n)) { 54 Build(1, n, 1); 55 scanf("%d", &q); 56 while (q--) { 57 scanf("%d%d", &x, &y); 58 printf("%d\n", Query(x, y, 1, n, 1).val); 59 } 60 } 61 return 0; 62 }