题干：

The shorter, the simpler. With this problem, you should be convinced of this truth. 
   
  You are given an array AA of NN postive integers, and MM queries in the form (l,r)(l,r). A function F(l,r) (1≤l≤r≤N)F(l,r) (1≤l≤r≤N) is defined as: 
F(l,r)={AlF(l,r−1) modArl=r;l<r.F(l,r)={All=r;F(l,r−1) modArl<r. 
You job is to calculate F(l,r)F(l,r), for each query (l,r)(l,r).

Input

There are multiple test cases. 
   
  The first line of input contains a integer TT, indicating number of test cases, and TT test cases follow. 
   
  For each test case, the first line contains an integer N(1≤N≤100000)N(1≤N≤100000). 
  The second line contains NN space-separated positive integers: A1,…,AN (0≤Ai≤109)A1,…,AN (0≤Ai≤109). 
  The third line contains an integer MM denoting the number of queries. 
  The following MM lines each contain two integers l,r (1≤l≤r≤N)l,r (1≤l≤r≤N), representing a query.

Output

For each query(l,r)(l,r), output F(l,r)F(l,r) on one line.

Sample Input

1
3
2 3 3
1
1 3

Sample Output

2

题目大意：

有n个数，m个查询，每个查询有一个区间[L, R], 求ans, ans = a[L]%a[L+1]%a[L+2]%...%a[R];

解题报告：

其实可以不用求区间[l,r]的，可以直接求后缀[l,n]的，可以简化代码。（比求区间的简单，并且肯定满足题意）这题还可以用RMQ+二分来代替线段树求解。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5; 
int a[MAX];
int n,q;
struct TREE {int l,r;int mi;
} tr[MAX<<2];
void pushup(int cur) {tr[cur].mi = min(tr[cur*2].mi,tr[cur*2+1].mi);
}
void build(int l,int r,int cur) {tr[cur].l = l,tr[cur].r = r;if(l == r) {tr[cur].mi = a[l]; return ;}int m = (l+r)>>1;build(l,m,cur*2);build(m+1,r,cur*2+1);pushup(cur);
}
int query(int pl,int pr,int val,int cur) {if(tr[cur].l == tr[cur].r) {if(tr[cur].mi <= val) return tr[cur].l;else return n+1;}if(pl <= tr[cur].l && pr >= tr[cur].r) {if(tr[cur*2].mi <= val) return query(pl,pr,val,cur*2);else if(tr[cur*2+1].mi <= val) return query(pl,pr,val,cur*2+1);else return n+1;}int res = n+1;if(pl <= tr[cur*2].r) res = query(pl,pr,val,cur*2);if(res != n+1) return res;if(pr >= tr[cur*2+1].l) return query(pl,pr,val,cur*2+1); else return n+1; 
}
int main()
{int T;cin>>T;while(T--) {scanf("%d",&n);for(int i = 1; i<=n; i++) scanf("%d",a+i);build(1,n,1);scanf("%d",&q);for(int l,r,i = 1; i<=q; i++) {scanf("%d%d",&l,&r);int ans = a[l];l++;while(l <= r && ans != 0) {int x = query(l,r,ans,1);if(x <= r) ans %= a[x];l = x+1;}printf("%d\n",ans);}		}return 0 ;
}