hdu 5726

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4549    Accepted Submission(s): 1630

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
  

 

Author

HIT

 

Source

2016 Multi-University Training Contest 1

思路：定义f[i][j]为：ai开始，连续2^j个数的最大公约数，令k=log2(r-l+1)，

         则look(l,r)=gcd(f[l][k],f[r-(1<<k)+1][k])(f[l][k] 和 f[r-(1<<k)+1][k]可能会有重叠，但不影响最终的gcd值。)

        对于第二问，枚举i（1-n），二分j， i-j都是同一个gcd值，然后直接加到map中。

#include<cstdio>
#include<cmath>
#include<map>
using namespace std;
int f[100010][18];
int a[100010];
int n,m;
int gcd(int a,int b){ return b?gcd(b,a%b):a;}
void rmq()
{
	for(int j=1;j<=n;j++) f[j][0]=a[j];
	for(int i=1;i<18;i++)
	{
		for(int j=1;j<=n;j++)
		  {
		  	if(j+(1<<i)-1<=n) f[j][i]=gcd(f[j][i-1],f[j+(1<<i-1)][i-1]);
		  }
	}
}
int look(int l,int r)
{
	int k=(int)log2((double)(r-l+1));
	return gcd(f[l][k],f[r-(1<<k)+1][k]);
}
map<int,long long>mp;
void ta()
{
	mp.clear();
	for(int i=1;i<=n;i++)
	  {
	  	int g=f[i][0],j=i;
	  	while(j<=n)
	  	  {
	  	  	int l=j,r=n;
	  	  	while(l<r)
	  	  	  {
	  	  	  	int mid=(l+r+1)>>1;
	  	  	  	if(look(i,mid)==g) l=mid;
	  	  	  	      else r=mid-1;
			  }
			mp[g]+=l-j+1;
			j=l+1;
			g=look(i,j);
		  }
	  }
}
int main()
{
	int t,l,r;
	int cas=1;
	scanf("%d",&t);
	while(t--)
	 {
	 	printf("Case #%d:\n",cas++);
	 	scanf("%d",&n);
	 	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	rmq();
	ta();
	scanf("%d",&m);
	for(int i=0;i<m;i++)
	{
		scanf("%d%d",&l,&r);
		int g=look(l,r);
		printf("%d %lld\n",g,mp[g]);
	}
	 }
}