大体题意：
告诉你一个自定义的F函数求解方法，给你q个自变量（区间），求解答案！
思路：
分析自定义的函数可以知道，我们可以变换一下这个函数，变换之后也就是F（l，r） = a[l] % a[l+1] %..... % a[r];
如果这样纯粹的求解的话，肯定会超时的，不说有多少个操作，就是这个取模就慢的不行！
我们知道，一个数对比他大的数取模的话，那么这个数取模后是不变的，我们可以利用这一个性质进行优化！
我们定义nest[i] 表示在i位置以后 第一个不比a[i]大的数，这样我们不断的跳位置即可！因为之间的没必要进行枚举，因为取模了值也不变！这样就可以通过了！
坑：
注意，输入的n 个数会有0存在，因此不能直接取模，如果发现是0的话，就没意义了， 直接break 即可！（continue就超时）反正有0就很恶心了，不该有0 的数据的 = = ！
详细见代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100000 + 10;
int n;
int nest[maxn];
int a[maxn];
void get_next(){
for (int i = 1; i <= n; ++i){
for (int j = i+1; j <= n; ++j){
nest[i] = n+1;
if (a[j] <= a[i]){
nest[i] = j;
break;
}
}
}
}
int main(){
int T;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for (int i = 1; i <= n; ++i){
scanf("%d",&a[i]);
}
get_next();
int q;
scanf("%d",&q);
while(q--){
int l,r;
scanf("%d %d",&l,&r);
int ans = a[l];
for (int i = nest[l]; i <= r; i = nest[i]){
if (a[i] == 0)break;
ans %= a[i];
if (ans == 0)break;
}
printf("%d\n",ans);
}
}
return 0;
}
Function
Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1548 Accepted Submission(s): 567
Problem Description
The shorter, the simpler. With this problem, you should be convinced of this truth.
You are given an array
A of
N postive integers, and
M queries in the form
(l,r). A function
F(l,r) (1≤l≤r≤N) is defined as:
F(l,r)={AlF(l,r−1) modArl=r;l<r.
You job is to calculate
F(l,r), for each query
(l,r).
Input
There are multiple test cases.
The first line of input contains a integer
T, indicating number of test cases, and
T test cases follow.
For each test case, the first line contains an integer
N(1≤N≤100000).
The second line contains
N space-separated positive integers:
A1,…,AN (0≤Ai≤109).
The third line contains an integer
M denoting the number of queries.
The following
M lines each contain two integers
l,r (1≤l≤r≤N), representing a query.
Output
(l,r), output F(l,r)
Sample Input
1 3 2 3 3 1 1 3
Sample Output
2
Source
2016 ACM/ICPC Asia Regional Dalian Online
