传送门

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description
Give you a sequence of $N(N≤100,000)$ integers : $a_1,\cdots,a_n(0<a_i≤1000,000,000)$. There are $Q(Q≤100,000)$ queries. For each query $l,r$ you have to calculate $\text{gcd}(a_l,,a_{l+1},\cdots,a_r)$ and count the number of pairs$(l′,r′)(1≤l<r≤N)$such that $\text{gcd}(a_{l′},a_{l′+1},\cdots,a_{r′})$ equal $\text{gcd}(a_l,a_{l+1},...,a_{r})$.

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, $a_1,\cdots,a_n(0<a_i≤1000,000,000)$.

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

For the next $Q$ lines, $i\text{-th}$ line contains two number , stand for the $l_i,r_i$, stand for the $i\text{-th}$ 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 $\text{gcd}(a_l,a_{l+1}, \cdots,a_r)$ and the second number stands for the number of pairs$(l′,r′)$ such that $\text{gcd}(a_{l′},a_{l′+1},\cdots,a_{r′})$ equal $\text{gcd}(a_l,a_{l+1},\cdots,a_r)$.

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

Author
HIT

Source
2016 Multi-University Training Contest 1

题意:

支持查询: (1) 区间gcd, (2) gcd值等于k的区间数

Solution:

区间gcd的查询线段树即可解决, 另外还能支持单点修改. 但这题要求支持查询gcd值等于k的区间个数, 线段树就有点乏力了, 因为这个信息大概不太好通过合并区间信息来得到. 我们来考虑区间gcd的性质:

令$\gcd_r(l)\quad (1\le l \le r) $表示, $l$到$r$的$\gcd$. 不难看出:

• $\gcd_r(l)$随着$l$的增大是单调不减的
• $\gcd_r(l)$最多取$\log{a_r}$个值, 因为在区间左端点从$r$移动到$l$的过程中gcd每缩小到一个新值都是因为除以了上个gcd的某个因子, 因而至少缩小为上个gcd的$\frac{1}{2}$,  从而不同的区间$\gcd$值最多有$\log{a_r}$个

因此, 我们可以对每个右端点$r$,  维护函数$\gcd_r(l)$. 实现方法是:

用vector<pair<int,int>> f 存某个函数$\gcd_r(l)$的每一段 (最多有$\log{a_r}$段), f[i].first表示第$i$段的左端点, f[i].second表示第$i$段的函数值.

在维护这$n$个函数的过程中, 用map记录每个$\gcd$出现的次数 (不同$\gcd$值最多有$O(n\log{N})$个, 实际上远达不到这个值.

接下来我们考虑如何利用上面维护好的函数查询某个区间$[l,r]$的$\gcd$.

我们可以在函数$g_r(l)$中二分查询小于等于的$l$的first的最大值对应的second的值, 这便是答案.

言不尽意, 详见代码.

Implementation:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3
 4 const int N(1e5+5);
 5 typedef pair<int,int> P;
 6
 7 vector<P> f[N];
 8 unordered_map<int,long long> cnt;
 9 int T, n, q, cs;
10
11 int main(){
12     for(cin>>T; T--; ){
13         cin>>n;
14         for(int i=1, gcd, pos; i<=n; i++){
15             scanf("%d", &gcd), pos=i, f[i].clear();
16             for(auto x:f[i-1]){
17                 if(__gcd(gcd, x.second)!=gcd) f[i].push_back({pos, gcd});
18                 gcd=__gcd(gcd, x.second), pos=x.first;
19             }
20             f[i].push_back({pos, gcd});
21         }
22         cnt.clear();
23         for(int i=1; i<=n; i++){
24             int pos=i+1;
25             for(auto x:f[i])
26                 cnt[x.second]+=pos-x.first, pos=x.first;
27         }
28         cin>>q;
29         printf("Case #%d:\n", ++cs);
30         for(int l, r; q--; ){
31             scanf("%d%d", &l, &r);
32             int gcd=lower_bound(f[r].begin(), f[r].end(), P(l, INT_MAX), greater<P>())->second;
33             printf("%d %lld\n", gcd, cnt[gcd]);
34         }
35     }
36 }