hdu5726 GCD(gcd +二分+rmq)

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：给你n个数，m个询问，每一个询问都是一个区间，让你先计算出这段区间所有数的gcd，然后问1~n所有连续区间中gcd的值等于询问区间gcd的区间个数。

思路：考虑到如果固定区间左端点L，那么右端点从L+1变化到n的过程中gcd最多变化log(区间内最大的数的大小)次（因为每次变化至少除以2）,那么我们就可以枚举左端点，然后每次二分值连续的区间，然后都存到map里就行了。

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