GCD

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
分析：因为固定左端点的GCD对右端点而言单调不增，所以可以二分。RMQ可以预处理所有区间的GCD;
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <ext/rope>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=1e5+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
using namespace __gnu_cxx;
int gcd(int p,int q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,a[maxn],p[maxn],st[20][maxn];
map<int,ll>ans;
void all_init()
{
    for(int i=2;i<=maxn-10;i++)p[i]=1+p[i/2];
}
void init()
{
    for(int i=1;i<20;i++)
        for(int j=1;j+(1<<i)-1<=n;j++)
         st[i][j]=gcd(st[i-1][j],st[i-1][j+(1<<(i-1))]);
}
int getgcd(int l,int r)
{
    int x=p[r-l+1];
    return gcd(st[x][l],st[x][r-(1<<x)+1]);
}
int work(int now,int l,int start)
{
    int pos,r=n;
    while(l<=r)
    {
        int mid=l+r>>1;
        if(getgcd(start,mid)>=now)
            pos=mid,l=mid+1;
        else r=mid-1;
    }
    return pos;
}
int main()
{
    int i,j,k,t;
    all_init();
    scanf("%d",&t);
    for(k=1;k<=t;k++)
    {
        ans.clear();
        scanf("%d",&n);
        for(i=1;i<=n;i++)scanf("%d",&st[0][i]);
        init();
        for(i=1;i<=n;i++)
        {
            for(j=i;j<=n;)
            {
                int q=getgcd(i,j),ending;
                ending=work(q,j,i);
                ans[q]+=ending-j+1;
                j=ending+1;
            }
        }
        printf("Case #%d:\n",k);
        scanf("%d",&m);
        while(m--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            j=getgcd(l,r);
            ll num=ans[j];
            printf("%d %lld\n",j,num);
        }
    }
    //system ("pause");
    return 0;
}