hdu5875 Function（暴力）

思路：预处理出右边第一个比它小的数，然后询问的话就跳跳跳就可以了....当然可以随便构造数据让它T...

#include<bits/stdc++.h>using namespace std;const int maxn = 100000+7;int pos[maxn];int a[maxn],n,m;void init(){memset(pos,-1,sizeof(pos));for(int i = n-1;i>=1;i--){        int p = i+1;for(;;){if(a[i]>a[p]){pos[i]=p;break;}if(pos[p]==-1){pos[i]=-1;break;}p = pos[p];}}}int main(){    int T;scanf("%d",&T);while(T--){scanf("%d",&n);for(int i = 1;i<=n;i++)scanf("%d",&a[i]);init();scanf("%d",&m);for(int i = 1;i<=m;i++){int x,y;scanf("%d%d",&x,&y);int ans = a[x];x = pos[x];while(x<=y && x!=-1){ans%=a[x];x=pos[x];}printf("%d\n",ans);}}}

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

132 3 311 3

 

Sample Output

2

 

Source

2016 ACM/ICPC Asia Regional Dalian Online