多校 GCD

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

151 2 4 6 741 52 43 44 4

 

Sample Output

Case #1:1 82 42 46 1

我们注意观察

gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​)，当l固定不动的时候，$r=l...n$时，我们可以容易的发现,随着$r$的増大，gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​)是递减的，同时gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​)最多 有$log1000,000,000$个不同的值，为什么呢？因为a_{l}a​l​​最多也就有$log1000,000,000$个质因数

所以我们可以在log级别的时间处理出所有的以L开头的左区间的gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​) 那么我们就可以在$nlog1000,000,000$的时间内预处理出所有的gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​)然后我们可以用一个map来记录，gcd值为key的有多少个 然后我们就可以对于每个询问只需要查询对应gcd(a_{l},a_{l+1},...,a_{r})gcd(a​l​​,a​l+1​​,...,a​r​​)为多少，然后再在map 里面查找对应答案即可.

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <queue>#include <vector>#include <stack>#include <map>#include <algorithm>#include <cmath>#include <ctime>#define LL long longusing namespace std;int a[110000];//pair.first为求的GCD//pair.second为区间的R//vec对应的下标为区间的Lvector<pair<int,int> >vec[110000];map<int,LL>M;int main(){    int T,t=0;    int l,r;    int n,m;    scanf("%d",&T);    while(T--){        scanf("%d",&n);        for(int i = 1; i <= n; ++i){            scanf("%d",&a[i]);            vec[i].clear();        }        M.clear();        for(int i = n; i >= 1; i--){            int g = 0;            /*            L+1....R3...R2....R1....R0            L+1....R3区间的GCD相同            L+1....R3+1和L...R3+2和...和L+1....R2的GCD相同            L+1....R2+1和L...R2+2和...和L+1....R1的GCD相同            L+1....R1+1和L...R1+2和...和L+1....R0的GCD相同            假设L+k<R3            __gcd(a[l],[L+1,L+k])和__gcd(a[l],[L+1,R3])相同            所以只需计算__gcd(a[l],[L+1,R3])的gcd就求出了[L,R3]的GCD了,他们相同。            以此类推...            */            for(int j = 0; j < vec[i+1].size(); ++j){                int x = vec[i+1][j].first;                int y = vec[i+1][j].second;                int d = __gcd(x,a[i]);                if(g==d)continue;                g = d;                vec[i].push_back(make_pair(g,y));            }            //[i,i]区间也是一个GCD            if(g!=a[i])vec[i].push_back(make_pair(a[i],i));            //用map记录gcd值出现了几次            for(int j = 0; j < vec[i].size(); ++j){                //区间[i,vec[i][j].second] 的 gcd 为 vec[i][j].first                if(j+1 != vec[i].size())                    M[vec[i][j].first] += (LL)(vec[i][j].second-vec[i][j+1].second);                else{                    M[vec[i][j].first] += (LL)(vec[i][j].second-i+1);                }            }        }        scanf("%d",&m);        printf("Case #%d:\n",++t);        while(m--){            scanf("%d%d",&l,&r);            int x = vec[l].size()-1;            for(int i = 0; i < vec[l].size(); ++i){                if(vec[l][i].second < r){                    x = i-1;                    break;                }            }            printf("%d %lld\n",vec[l][x].first,M[vec[l][x].first]);        }    }    return 0;}