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 integersa and b that is in a given range fromlow tohigh (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 2731 510 119 11

Output

3-19AC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;}