hdu5726 GCD （线段树+区间gcd）
GCD
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1587 Accepted Submission(s): 515
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.
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).
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
题意：给你n个数，然后给你一对l,r，求这n个数的子区间中gcd值等于gcd(l,l+1,...,r)的个数
思路：首先我们考虑求出一个区间的gcd，这里可以用线段树求出，主要的就是查询符合条件的子区间的个数，这里需要预处理所有子区间了，小于N的子区间的gcd种类数最多为logN个，所以说我们用map存储gcd的个数和种类，以a[i]为右端点进行扫描，右端点不断向后移动，以a[i]为右端点的gcd=gcd(以a[i-1]为右端点的gcd，a[i])，我们通过记录以前一个元素为右端点的gcd，然后扫描就可以了，因为最多会有logn个，这样就可以求出所有子区间的gcd种类和个数，最后查询的时候O(1)的复杂度，总的复杂度O(nlognlogn)。
ac代码：
Case #1: 1 8 2 4 2 4 6 1
题意：给你n个数，然后给你一对l,r，求这n个数的子区间中gcd值等于gcd(l,l+1,...,r)的个数
思路：首先我们考虑求出一个区间的gcd，这里可以用线段树求出，主要的就是查询符合条件的子区间的个数，这里需要预处理所有子区间了，小于N的子区间的gcd种类数最多为logN个，所以说我们用map存储gcd的个数和种类，以a[i]为右端点进行扫描，右端点不断向后移动，以a[i]为右端点的gcd=gcd(以a[i-1]为右端点的gcd，a[i])，我们通过记录以前一个元素为右端点的gcd，然后扫描就可以了，因为最多会有logn个，这样就可以求出所有子区间的gcd种类和个数，最后查询的时候O(1)的复杂度，总的复杂度O(nlognlogn)。
ac代码：
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