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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