HDU 5726 GCD

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 $ ext{gcd}(a_l,,a_{l+1},cdots,a_r)$ and count the number of pairs$(l′,r′)(1≤l<r≤N)$such that $ ext{gcd}(a_{l′},a_{l′+1},cdots,a_{r′})$ equal $ ext{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, $i ext{-th}$ line contains two number , stand for the $l_i,r_i$, stand for the $i ext{-th}$ 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 $ ext{gcd}(a_l,a_{l+1}, cdots,a_r)$ and the second number stands for the number of pairs$(l′,r′)$ such that $ ext{gcd}(a_{l′},a_{l′+1},cdots,a_{r′})$ equal $ ext{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
 

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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 (1le 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$O(nlog{N})$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:

#include <bits/stdc++.h>
using namespace std;

const int N(1e5+5);
typedef pair<int,int> P;

vector<P> f[N];
unordered_map<int,long long> cnt;
int T, n, q, cs;

int main(){
    for(cin>>T; T--; ){
        cin>>n;
        for(int i=1, gcd, pos; i<=n; i++){
            scanf("%d", &gcd), pos=i, f[i].clear();
            for(auto x:f[i-1]){
                if(__gcd(gcd, x.second)!=gcd) f[i].push_back({pos, gcd});
                gcd=__gcd(gcd, x.second), pos=x.first;        
            }
            f[i].push_back({pos, gcd});
        }
        cnt.clear();
        for(int i=1; i<=n; i++){
            int pos=i+1;
            for(auto x:f[i])
                cnt[x.second]+=pos-x.first, pos=x.first;
        }
        cin>>q;
        printf("Case #%d:
", ++cs);
        for(int l, r; q--; ){
            scanf("%d%d", &l, &r);
            int gcd=lower_bound(f[r].begin(), f[r].end(), P(l, INT_MAX), greater<P>())->second;
            printf("%d %lld
", gcd, cnt[gcd]);
        }
    }
}