hdu 3074 Multiply game

Problem Description

Tired of playing computer games, alpc23 is planning to play a game on numbers. Because plus and subtraction is too easy for this gay, he wants to do some multiplication in a number sequence. After playing it a few times, he has found it is also too boring. So he plan to do a more challenge job: he wants to change several numbers in this sequence and also work out the multiplication of all the number in a subsequence of the whole sequence.
  To be a friend of this gay, you have been invented by him to play this interesting game with him. Of course, you need to work out the answers faster than him to get a free lunch, He he…

Input

The first line is the number of case T (T<=10).
  For each test case, the first line is the length of sequence n (n<=50000), the second line has n numbers, they are the initial n numbers of the sequence a1,a2, …,an, 
Then the third line is the number of operation q (q<=50000), from the fourth line to the q+3 line are the description of the q operations. They are the one of the two forms:
0 k1 k2; you need to work out the multiplication of the subsequence from k1 to k2, inclusive. (1<=k1<=k2<=n) 
1 k p; the kth number of the sequence has been change to p. (1<=k<=n)
You can assume that all the numbers before and after the replacement are no larger than 1 million.

Output

For each of the first operation, you need to output the answer of multiplication in each line, because the answer can be very large, so can only output the answer after mod 1000000007.

Sample Input

161 2 4 5 6 330 2 51 3 70 2 5

Sample Output

240
420
这个题，一道线段树的题，题目的意思容易理解。就是
先输入一个序列，然后 后面m个数据。 第一个为0的话，就
输出第k1 到k2的之间的积，如果是0的话，就把第k 的值改成p
注意积的值对100000007取模， long long 下  lld输出即可
无语的wa点在定义树的结构体的时候，没把节点的值long long
找错 那个心累
#include<iostream>#include<cstring>using namespace std;const int mod=1000000007;struct node{int left,right;long long num;}s[150005];int a[50010];void build(int ll,int rr,int i) //建线段树 {int mm;s[i].left=ll;  s[i].right=rr;   //左右区间给树的数组 if(ll==rr)   //达到叶结点 {s[i].num=a[ll]; //先用数组存下该队的人return ;}mm=(ll+rr)/2;  //二分思想build(ll,mm,2*i);  //左子树build(mm+1,rr,2*i+1); //右子树s[i].num=(s[2*i].num*s[2*i+1].num)%mod;}void updata(int y,int x,int i) //值   位置  {if(s[i].left==x&&x==s[i].right) {s[i].num=y;return; //到达叶子节点} if(x<=s[2*i].right)  updata(y,x,2*i);if(x>=s[2*i+1].left) updata(y,x,2*i+1);s[i].num=(s[2*i].num*s[2*i+1].num)%mod;}long long sum(int ll,int rr,int i) //查询 {int mm;if(s[i].left==ll&&rr==s[i].right) //如果区间覆盖，则直接返回其节点的值 {return s[i].num;}mm=(s[i].left+s[i].right)/2;if(rr<=mm)  return sum(ll,rr,2*i); //左子树 if(ll>mm) return sum(ll,rr,2*i+1); //右子树 else {long long a=sum(ll,mm,2*i);long long b=sum(mm+1,rr,2*i+1);return (a*b)%mod; //左右之间 }}int main(){int bbs,k,i,j,n,m,x,y;scanf("%d",&bbs);while(bbs--){scanf("%d",&n);for(i=1;i<=n;i++)scanf("%d",&a[i]);build(1,n,1); //建树scanf("%d",&m);while(m--){scanf("%d%d%d",&k,&x,&y);if(k==0){long long ans=sum(x,y,1);ans=ans%mod;printf("%lld\n",ans);}else updata(y,x,1);  //值 和 位置  } }return 0;}