SPOJ 1043 Can you answer these queries I 求任意区间最大连续子段和 线段树

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维护区间左起连续的最大和，右起连续的和。。

#include <cstdio>  #include <iostream>  #include <algorithm>  #include <string.h>  #include <math.h>  #include <vector>  #include <map>  using namespace std;  #define N 50050  #define Lson(x) tree[x].l  #define Rson(x) tree[x].r  #define L(x) (x<<1)  #define R(x) (x<<1|1)  #define Sum(x) tree[x].sum  #define Max(x) tree[x].max  #define Lmax(x) tree[x].lmax  #define Rmax(x) tree[x].rmax  struct node{      int l, r;      int mid(){return (l+r)>>1;}      int lmax, rmax, max, sum;  }tree[N<<2];  int n, a[N], Q;  void push_down(int id){}  void push_up(int id){      Lmax(id) = max(Lmax(L(id)), Sum(L(id)) + Lmax(R(id)));      Rmax(id) = max(Rmax(R(id)), Sum(R(id)) + Rmax(L(id)));      Sum(id) = Sum(L(id)) + Sum(R(id));      Max(id) = max(max(Max(L(id)), Max(R(id))), Rmax(L(id)) + Lmax(R(id)));  }  void updata_point(int val, int id){Lmax(id) = Rmax(id) = Max(id) = Sum(id) = val;}  void build(int l, int r, int id){      Lson(id) = l; Rson(id) = r;      if(l == r)      {           updata_point(a[l], id);          return;      }      int mid = tree[id].mid();      build(l, mid, L(id));      build(mid+1, r, R(id));      push_up(id);  }  void updata(int pos, int val, int id){      push_down(id);      if(Lson(id) == Rson(id))      {          updata_point(val, id);          return ;      }      int mid = tree[id].mid();      if(mid < pos)          updata(pos, val, R(id));      else           updata(pos, val, L(id));  }  int query_l(int l, int r, int id){      push_down(id);      if(l == Lson(id) && Rson(id) == r) return Lmax(id);      int mid = tree[id].mid();      if(mid < l)          return query_l(l, r, R(id));      else if(r <= mid)          return query_l(l, r, L(id));      int lans = query_l(l, mid, L(id)), rans = query_l(mid+1, r, R(id));      return max(lans, Sum(L(id)) + rans);  }  int query_r(int l, int r, int id){      push_down(id);      if(l == Lson(id) && Rson(id) == r) return Rmax(id);      int mid = tree[id].mid();      if(mid < l)          return query_r(l, r, R(id));      else if(r <= mid)          return query_r(l, r, L(id));      int lans = query_r(l, mid, L(id)), rans = query_r(mid+1, r, R(id));      return max(rans, Sum(R(id)) + lans);  }  int query(int l, int r, int id){      push_down(id);      if(l == Lson(id) && Rson(id) == r)return Max(id);      int mid = tree[id].mid();      if(mid < l)          return query(l, r, R(id));      else if(r<=mid)          return query(l, r, L(id));      int lans = query(l, mid, L(id)), rans = query(mid+1, r, R(id));      int ans = max(lans, rans);      return max(ans, query_r(l, mid, L(id)) + query_l(mid+1, r, R(id)));  }  int main(){      while(~scanf("%d",&n)){          for(int i = 1; i <= n; i++)scanf("%d",&a[i]);          build(1, n, 1);          scanf("%d",&Q);          while(Q--){              int l, r;              scanf("%d %d",&l,&r);              printf("%d\n", query(l, r, 1));          }      }      return 0;  }  /* 3  -1 2 3 1 1 2  */