Function

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1866    Accepted Submission(s): 674

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

这个题目完全可以出个单调递减的序列卡时间啊...不知道时间复杂度怎么降下来的..因为右边比当前数大的数字是造不成影响的,所以我们用单调栈预处理出每个的右边,这样就可以跳着找了..但是我觉得数据强点不靠谱啊..

///pro do this : a[l]%a[l+1]%...%a[r]
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <string.h>
#include <vector>
using namespace std;
typedef long long LL;
const LL INF = 1e10;
const int N = 100005;
LL a[N],R[N];
int main(){
    int tcase,n;
    scanf("%d",&tcase);
    while(tcase--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i]);
        }
        memset(R,-1,sizeof(R));
        for(int i=n-1;i>=1;i--){ ///单调栈维护其右边小于 a[i] 的第一个数
            int t =i+1;
            while(true){
                if(a[i]>=a[t]){
                    R[i] = t;
                    break;
                }
                if(R[t]==-1){
                    break;
                }
                t = R[t];
            }
            R[i] = t;
        }
        int m;
        scanf("%d",&m);
        while(m--){
            int l,r;
            scanf("%d%d",&l,&r);
            LL ans = a[l];
            int nxt = l;
            while(R[nxt]<=r&&R[nxt]!=-1){
                nxt = R[nxt];
                ans = ans%a[nxt];
            }
            printf("%lld\n",ans);
        }
    }
    return 0;
}