C. Modified GCD(二分加搜索约数)

C. Modified GCD

time limit per test  2 seconds

memory limit per test   256 megabytes

题目连接: 传送门

Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.

A common divisor for two positive numbers is a number which both numbers are divisible by.

But your teacher wants to give you a harder task, in this task you have to find the greatest common divisord between two integers a and b that is in a given range fromlow to high (inclusive), i.e.low ≤ d ≤ high. It is possible that there is no common divisor in the given range.

You will be given the two integers a and b, then n queries. Each query is a range from low tohigh and you have to answer each query.

Input

The first line contains two integers a and b, the two integers as described above (1 ≤ a, b ≤ 10⁹). The second line contains one integern, the number of queries (1 ≤ n ≤ 10⁴). Thenn lines follow, each line contains one query consisting of two integers,low and high (1 ≤ low ≤ high ≤ 10⁹).

Output

Print n lines. Thei-th of them should contain the result of thei-th query in the input. If there is no common divisor in the given range for any query, you should print-1 as a result for this query.

Sample test(s)

Input

9 27
3
1 5
10 11
9 11

Output

3
-1
9

AC代码:

#include <algorithm>
#include <iostream>
#include <cstdio>

using namespace std;

const int M = 1e6;
int ma[M];

int Gcd(int a, int b)
{
    return b == 0 ? a : Gcd(b, a % b);
}

int main()
{
    int a,b,n,ua,ub;
    while(~scanf("%d %d",&a,&b))
    {
        scanf("%d",&n);
        int ans = Gcd(a,b), to = 0;
        for(int i = 1; i * i <= ans; i++) //运用求素数的方法sqrt时间复杂度找出所有约数;
        {
            if(ans % i == 0)
            {
                ma[to++] = i;
                if(i * i != ans) ma[to++] = ans / i;
            }
        }
        sort(ma,ma + to);
        while(n--)
        {
            scanf("%d %d",&ua,&ub);
            int tp = lower_bound(ma, ma + to, ub) - ma;
            if(ma[tp] > ub || tp == to) tp--;
            if(ma[tp] < ua || ma[tp] > ub) puts("-1");
            else printf("%d\n",ma[tp]);
        }
    }
    return 0;
}