hdu-3074 Multiply game---线段树+单点更新

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=3074

题目大意：

给一些数，进行点更新或者是区间计算乘积

解题思路：

裸的线段树，注意空间开大点

 #include<bits/stdc++.h>
 #define MID(l, r) ((l) + ((r) - (l)) / 2)
 #define lson(o) ((o)<<1)
 #define rson(o) ((o)<<1|1)
 using namespace std;
 typedef long long ll;
 const int maxn = 1e6 + ;
 const int mod = 1e9 + ;
 struct node
 {
     ll l, r, sum;
 }tree[maxn];
 ll a[maxn];
 void build(int o, int l, int r)
 {
     tree[o].l = l, tree[o].r = r;
     if(l == r)
     {
         tree[o].sum = a[l];
         return;
     }
     int m = MID(l, r);
     int lc = lson(o), rc = rson(o);
     build(lc, l, m);
     build(rc, m + , r);
     tree[o].sum = tree[lc].sum * tree[rc].sum % mod;
 }

 int ql, qr;//查询区间[ql, qr]中的max,min,sum
 ll ans_sum;
 void query(int o)
 {
     if(ql <= tree[o].l && qr >= tree[o].r)//[L, R]包含在[ql, qr]区间内，直接用该节点的信息，达到线段树查询快的操作
     {
         ans_sum *= tree[o].sum;
         ans_sum %= mod;
         return;
     }
     int m = MID(tree[o].l, tree[o].r);
     if(ql <= m)query(lson(o));
     if(qr > m)query(rson(o));
 }

 //单点更新，a[p] += v;
 int p, v;
 void update(int o)
 {
     if(tree[o].l == tree[o].r)
     {
         tree[o].sum = v;
         return;
     }
     int m = MID(tree[o].l, tree[o].r);
     int lc = lson(o), rc = rson(o);
     if(p <= m)update(lc);
     else update(rc);
     tree[o].sum = tree[lc].sum * tree[rc].sum % mod;
 }

 int main()
 {
     int T;
     scanf("%d", &T);
     while(T--)
     {
         int n;
         scanf("%d", &n);
         for(int i = ; i <= n; i++)scanf("%lld", &a[i]);
         build(, , n);
         int q, f, l, r;
         scanf("%d", &q);
         while(q--)
         {
             scanf("%d%d%d", &f, &l, &r);
             if(f)
             {
                 p = l, v = r;
                 update();
             }
             else
             {
                 ql = l, qr = r;
                 ans_sum = ;
                 query();
                 printf("%lld\n", ans_sum);
             }
         }
     }
     return ;
 }