HDU 5726 GCD(RMQ + 二分)

GCD

Problem Description

Give you a sequence of N(N≤100,000) integers : a1,...,an(0<ai≤1000,000,000). There are Q(Q≤100,000) queries. For each query l,r you have to calculate gcd(al,,al+1,...,ar) and count the number of pairs (l′,r′)(1≤l<r≤N)such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).

 

Input

The first line of input contains a number T, which stands for the number of test cases you need to solve.

The first line of each case contains a number N, denoting the number of integers.

The second line contains N integers, a1,...,an(0<ai≤1000,000,000).

The third line contains a number Q, denoting the number of queries.

For the next Q lines, i-th line contains two number , stand for the li,ri, stand for the i-th queries.

 

Output

For each case, you need to output “Case #:t” at the beginning.(with quotes, t means the number of the test case, begin from 1).

For each query, you need to output the two numbers in a line. The first number stands for gcd(al,al+1,...,ar) and the second number stands for the number of pairs (l′,r′) such that gcd(al′,al′+1,...,ar′) equal gcd(al,al+1,...,ar).

 

Sample Input

1 5 1 2 4 6 7 4 1 5 2 4 3 4 4 4

 

Sample Output

Case #1: 1 8 2 4 2 4 6 1 题意：给出一个数列 ，m次询问，每次询问 l，r区间内的gcd值 和 与该区间gcd值相同的区间有多少个 分析： 枚举每一个左端点，找每个左端点对应的所有gcd值区间，预处理出来，由于gcd值呈阶梯下降，所以完全可以处理，此时顺便用map统计区间个数 一开始考虑的是用线段树取gcd值，在加上二分处理，但是发现这样做其实复杂度达到了 n*logn*logn*logn ，直接T掉，然后想到用RMQ O（1）去处理gcd值，降下一个logn成功AC

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