HDU5875

Function

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 399    Accepted Submission(s): 151

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

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

 

Sample Input

1

3

2 3 3

1

1 3

 

Sample Output

2

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

 

 //2016.9.11
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #define N 100005

 using namespace std;

 int a[N], nex[N];//nex数组，表示跳到下一个要取余的位置，比a[i]大的数不用取余，此处优化降低时间

 int main()
 {
     int T, n, q, ans;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%d", &n);
         for(int i = ; i <= n; i++)
         {
             scanf("%d", &a[i]);
         }
         scanf("%d", &q);
         int l, r;
         for(int i = ; i <= n; i++)
         {
             nex[i] = -;
             for(int j = i+; j <= n; j++)
                 if(a[i]>=a[j])
                 {
                     nex[i] = j;
                     break;
                 }
         }
         while(q--)
         {
             scanf("%d%d", &l, &r);
             ans = a[l];
             for(int i = nex[l]; i <= r; i = nex[i])
             {
                 if(i == -)break;
                 ans %= a[i];
             }
             printf("%d\n", ans);
         }
     }

     return ;
 }