【HDU 5726】GCD（映射+RMQ）

【HDU 5726】GCD（映射+RMQ）

GCD

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1271    Accepted Submission(s): 391

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

 

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wange2014   |   We have carefully selected several similar problems for you:   5733  5732  5731  5730  5729 

题目大意：

给出一个长为n的数列a，0 < ai <= 1e9

Q次查询，每次询问区间[L,R]的GCD，还有整个数列中同样GCD的区间个数

问题一，用RMQ即可搞定，因为GCD跟Min是一种极性，都是不增函数

问题二，暴力统计一下即可，从a[0]到a[n]，一个映射存放全局的各GCD区间个数，一个映射存放遍历到当前位置，连续的各GCD区间个数，这样用当前的a[i]与连续到当前未知的各GCD求gcd，然后累加到全局的映射中，同时建立新的更新的映射，同时把a[i]加入新映射，表示a[i]自己这个区间

代码如下：

#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread(ch) freopen(ch,"r",stdin)
#define fwrite(ch) freopen(ch,"w",stdout)

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;

map <int,LL> tmp;
map <int,LL> a,b;
map <int,LL>::iterator iter;
int rmq[111111][22];
int cs[111111];

int Search(int l,int r)
{
	int cnt = (r-l+1);
	int k = cs[cnt];
	//printf("%d %d %d %d %d\n",l,r-cnt+1,rmq[l][k],rmq[r-cnt+1][k],k);
	return __gcd(rmq[l][k],rmq[r-(1<<k)+1][k]);
}

void init(int n)
{
	int tot = log(n*1.0)/log(2.0);
	int j = -1;
	int x = 1;
	cs[0] = 0;
	b.clear();
	a.clear();

	for(int k = 0; k <= tot; ++k)
	{
		for(int i = 0; i < n && (i+(1<<k)) <= n; ++i)
		{
			if(!k) 
			{
				scanf("%d",&rmq[i][k]);
				if(i+1 == x)
				{
					x <<= 1;
					j++;
				}
				cs[i+1] = j;

				//int ssss = log((i+1)*1.0)/log(2.0);
				//printf("%d %d %d\n",i+1,cs[i+1],ssss);
	
	
				tmp.clear();
				for(iter = a.begin(); iter != a.end(); ++iter)
				{
					int g = __gcd(iter->first,rmq[i][0]);
					tmp[g] += iter->second;
				}
				tmp[rmq[i][0]]++;
				a = tmp;
				for(iter = a.begin(); iter != a.end(); ++iter)
					b[iter->first] += iter->second;
			}
			else rmq[i][k] = __gcd(rmq[i][k-1],rmq[i+(1<<(k-1))][k-1]);
		}
	}
}

int main()
{
	//fread("");
	//fwrite("");

	int t,n;
	int z = 1;

	scanf("%d",&t);

	while(t--)
	{
		printf("Case #%d:\n",z++);
		scanf("%d",&n);
		init(n);

		int q;
		scanf("%d",&q);

		int l,r;
		while(q--)
		{
			scanf("%d%d",&l,&r);
			int g = Search(l-1,r-1);
			printf("%d %lld\n",g,b[g]);
		}
	}

	return 0;
}