GCD(st表+二分)

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3432    Accepted Submission(s): 1227

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

 

 

题意:第一行T代表测试组数,第二行一个整数 N 代表长为 N 的序列，再一行是一个整数 Q ，代表有 Q 次询问，

然后 Q 行,每行两个整数 l,r 输出区间[l,r]的gcd 和有多少个区间 gcd 等于 gcd[l,r]

 

题解:想了很久，没想到好办法，后来发现RMQ这种问题，线段树还是不行的，如果离线的话，st表查询时间复杂度比线段树更小，为O(1)

st算法是用来求解给定区间RMQ的最值

st表学习:http://www.cnblogs.com/yangjiyuan/p/5493274.html

学了st表后，查询有多少区间gcd==gcd[l,r]还是不知道怎么搞，题解是,用二分离线预处理...

 #include<iostream>
 #include<cstdlib>
 #include<cstring>
 #include<cstdio>
 #include<map>
 using namespace std;

 #define LL long long
 #define MAXN 100005

 int n;
 int num[MAXN];
 int mn[MAXN];
 int dp[MAXN][];
 map<int,LL> res;

 int gcd(int a,int b)
 {   return b==?a:gcd(b,a%b);}

 void Init()
 {
     mn[]=-;
     for (int i=;i<=n;i++)
     {
         mn[i]=((i&(i-))==)?mn[i-]+:mn[i-];
         dp[i][]=num[i];
     }
     for (int j=;j<=mn[n];j++)
         for (int i=;i+(<<j)-<=n;i++)
         {
             dp[i][j]=gcd(dp[i][j-],dp[i+(<<(j-))][j-]);
         }
 }

 int st_calc(int l,int r)
 {
     int k = mn[r-l+];
     return gcd(dp[l][k],dp[r-(<<k)+][k]);
 }

 void erfen()
 {
     res.clear();
     for (int i=;i<=n;i++)  //以i为左起点的区间的所有公约数
     {
         int cur = i , gc = num[i];  //用二分求出
         while (cur <= n)
         {
             int l=cur,r=n;
             while(l<r)
             {
                 int mid = (l+r+)>>;
                 if (st_calc(i,mid)==gc) l=mid;
                 else r = mid -;
             }
             if (res.count(gc)) res[gc]+=l-cur+;
             else res[gc]=l-cur+;
             cur = l + ;
             if (cur<=n)
                 gc = gcd(gc,num[cur]);
         }
     }
 }

 int main()
 {
     int t;
     int cas=;
     scanf("%d",&t);
     while (t--)
     {
         scanf("%d",&n);
         for (int i=;i<=n;i++)
             scanf("%d",&num[i]);
         Init();

         erfen();

         int q;
         scanf("%d",&q);
         printf("Case #%d:\n",++cas);
         while (q--)
         {
             int x,y;
             scanf("%d%d",&x,&y);
             int kk = st_calc(x,y);
             printf("%d %lld\n",kk,res[kk]);
         }
     }
     return ;
 }