2016暑假多校联合---GCD

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

思路：我们注意观察gcd(a​l​​,a​l+1​​,...,a​r​​)，当l固定不动的时候，r=l...nr=l...n时，我们可以容易的发现,随着rr的増大，gcd(a​l​​,a​l+1​​,...,a​r​​)是递减的，同时gcd(a​l​​,a​l+1​​,...,a​r​​)最多 有log 1000,000,000个不同的值，为什么呢？因为a​l​​最多也就有log 1000,000,000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​​) 那么我们就可以在n log 1000,000,000的时间内预处理出所有的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​​)为多少，然后再在map 里面查找对应答案即可.

 代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <map>
#include <vector>
using namespace std;
int a[100005];

vector<pair<int,int> > v[100005];
map<int,long long>ans;

int __gcd(int x,int y)
{
    int r=x%y;
    x=y;
    y=r;
    if(r==0) return x;
    return __gcd(x,y);
}

int main()
{
    int T,Case=0;
    int n;
    cin>>T;
    while(T--)
    {
        ans.clear();
        cin>>n;
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
        {
            int tot=0;
            for(int j=0;j<v[i-1].size();j++)
            {
                int s1=v[i-1][j].first;
                int s2=v[i-1][j].second;
                int  r=__gcd(a[i],s1);
                if(tot==r) continue;
                tot=r;
                v[i].push_back(make_pair(r,s2));
            }
            if(tot!=a[i])   v[i].push_back(make_pair(a[i],i));
            for(int j=0;j<v[i].size();j++)
            {
                if(j+1==v[i].size())
                    ans[v[i][j].first]+=i+1-v[i][j].second;
                else
                    ans[v[i][j].first]+=v[i][j+1].second-v[i][j].second;
            }
        }
        cout<<"Case #"<<(++Case)<<":"<<endl;
        int Q;
        cin>>Q;
        while(Q--)
        {
            int i,l,r;
            scanf("%d%d",&l,&r);
            for(i=0;i<v[r].size();i++)
            {
                if(v[r][i].second>l) break;
            }
            printf("%d %I64d\n",v[r][i-1].first,ans[v[r][i-1].first]);
        }
        for(int i=0;i<100005;i++)
            v[i].clear();
    }
    return 0;
}