HDU5726(RMQ&&二分)

GCD

Time Limit:5000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

//2016.8.9
#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
#include<cmath>

using namespace std;

typedef long long ll;
const int N = 100005;
int dp[N][20];//d[i][j]表示从第i个数字开始向后2^j个数字这段区间内的gcd,具有递减性
map<int, ll> mp;

void init_rmq(int n)//初始化dp,求出每段区间的gcd
{
    for(int j = 1; j < (int)log2(n)+1; j++)
      for(int i = 1; i <= n; i++)
      {
          if(i+(1<<j)-1 <= n)
            dp[i][j] = __gcd(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
      }
}

int rmq(int l, int r)//查询
{
    int k = (int)log2(r-l+1);
    return __gcd(dp[l][k], dp[r-(1<<k)+1][k]);
}

int main()
{
    int n, q, l, r, T, kase = 0;
    cin>>T;
    while(T--)
    {
        printf("Case #%d:
", ++kase);
        cin>>n;
        mp.clear();
        for(int i = 1; i <= n; i++)
          scanf("%d", &dp[i][0]);
        init_rmq(n);

//利用二分求具有相同gcd区间的数目
//-----------------------------------------------------------------------------------
        for(int i = 1; i <= n; i++)
        {
            int a = i, b = n, mid, tmp, vs;
            while(1)
            {
                tmp = a; 
                vs = rmq(i, a);
                while(a <= b)
                {
                    mid = (a+b)>>1;
                    if(rmq(i, mid)<vs) b = mid-1;
                    else a = mid+1;
                }
                mp[vs]+=1ll*(b-tmp+1);
                b = n;
                if(a>b)break;
            }
        }
//------------------------------------------------------------------------------------

        cin>>q;
        while(q--)
        {
            scanf("%d%d", &l, &r);
            int ans = rmq(l, r);
            cout<<ans<<" "<<mp[ans]<<endl;
        }
    }

    return 0;
}