hdu-5726 GCD(rmq)

题目链接：

GCD

Time Limit: 10000/5000 MS (Java/Others)   

 Memory Limit: 65536/65536 K (Java/Others)

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：

 

给含有n个数的序列,现在询问[l,r]的gcd是多少,同时还要求与gcd相同的区间有多少个;

 

思路:

 

静态区间可以用rmq,先预处理就可以O(1)的回答询问了;还有就是询问区间的个数,可以先预处理出所有的gcd,用map存下来,预处理的时候枚举左端点,由于gcd是单调递减的,所以可以分层后二分右端点,找到每层的长度,加到map中,由于一个数的因子个数不超过log(n)所以层数比较少才可以这样处理,写的线段树t了,看来静态区间查询还是rmq好;

 

AC代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e5+10;
const int maxn=500+10;
const double eps=1e-10;

int a[N],n,d[N][20];
map<int,LL>mp;

int gcd(int x,int y)
{
	if(y==0)return x;
	return gcd(y,x%y);
}

inline void rmq()
{
	for(int i=1;i<=n;i++)d[i][0]=a[i];
	for(int j=1;(1<<j)<=n;j++)
	{
		for(int i=1;i+(1<<j)-1<=n;i++)
		{
			d[i][j]=gcd(d[i][j-1],d[i+(1<<(j-1))][j-1]);
		}
	}
	return ;
}
inline int query(int l,int r)
{
	int k=0;
	while((1<<(k+1))<=r-l+1)k++;
	return gcd(d[l][k],d[r-(1<<k)+1][k]);
}
inline void Init()
{
	mp.clear();
	For(i,1,n)
	{
		int pos=i;
		while(pos<=n)
		{
			int le=query(i,pos);
			int l=pos,r=n;
			while(l<=r)
			{
				int mid=(l+r)>>1;
				if(query(i,mid)<le)r=mid-1;
				else l=mid+1;
			}
			mp[le]+=(LL)(r-pos+1);
			pos=r+1;
		}
	}
}
int main()
{
		int t,Case=0;
		read(t);
		while(t--)
		{
			printf("Case #%d:
",++Case);
			read(n);
			For(i,1,n)read(a[i]);
			rmq();
			Init();
			int q;
			read(q);
			int l,r;
			while(q--)
			{
				read(l);read(r);
				int ans=query(l,r);
				printf("%d ",ans);
				print(mp[ans]);
			}
		}

        return 0;
}