HDU5726-GCD

GCD

                                                                          Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                                       Total Submission(s): 3629    Accepted Submission(s): 1305

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

 

Recommend

wange2014

 

题意：给出n个数，有q次询问，每个给出一个区间，问有多少个区间的gcd和这个区间的gcd相同

解题思路：先预处理出每个位置为左端点，通过二分+RMQ找出向右每种gcd有几个，用map保存每种gcd有几个

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int dp[100009][20],a[100009],n,q,x;

int gcd(int a, int b)
{
	if (a > b) swap(a, b);
	while (b%a)
	{
		int k = b%a;
		b = a;
		a = k;
	}
	return a;
}

void init()
{
	for (int i = 1; i <= n; i++) dp[i][0] = a[i];
	for (int i = 1; (1 << i) <= n; i++)
		for (int j = 1; j + (1 << i) - 1 <= n; j++)
			dp[j][i] = gcd(dp[j][i - 1], dp[j + (1 << (i - 1))][i - 1]);
}

int query(int l, int r)
{
	int k = 0;
	while (r - l + 1 >= 1 << (k + 1)) k++;
	return gcd(dp[l][k], dp[r - (1 << k) + 1][k]);
}

int solve(int a, int p,int pp)
{
	int l = pp, r = n,ans;
	while (l <= r)
	{
		int mid = (l + r) >> 1;
		int k = query(p, mid);
		if (k == x) { ans = mid, l = mid + 1; }
		else if (k < x) r = mid - 1;
		else l = mid + 1;
	}
	return ans;
}

int main()
{
	int t, cas = 0;
	scanf("%d", &t);
	while (t--)
	{
		printf("Case #%d:\n", ++cas);
		scanf("%d", &n);
		for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
		init();
		map<int, LL>mp;
		for (int i = 1; i <= n; i++)
		{
			x = a[i];
			int pos = i;
			while (pos<=n)
			{
				int pos1 = solve(x,i, pos)+1;
				mp[x] += (pos1 - pos);
				if (pos1 > n) break;
				x = query(i,pos1);
				pos = pos1;
			}
		}
		scanf("%d", &q);
		while (q--)
		{
			int l, r;
			scanf("%d%d", &l, &r);
			x = query(l, r);
			printf("%d %lld\n", x, mp[x]);
		}
	}
	return 0;
}