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 = ;
 int dp[N][];//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 = ; j < (int)log2(n)+; j++)
       for(int i = ; i <= n; i++)
       {
           if(i+(<<j)- <= n)
             dp[i][j] = __gcd(dp[i][j-], dp[i+(<<(j-))][j-]);
       }
 }

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

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

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