I - Modified GCD （最大公约数，打表）

最新推荐文章于 2020-07-27 12:24:00 发布

philophobia

最新推荐文章于 2020-07-27 12:24:00 发布

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本文介绍了一种算法，该算法首先计算两个给定整数的最大公约数(GCD)，接着找出在特定区间内存在的公约数。通过枚举并筛选最大公约数的因数来解决这一问题。

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 divisor d between two integers a and b that is in a given range from low 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 to high 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 integer n, the number of queries (1 ≤ n ≤ 104). Then n lines follow, each line contains one query consisting of two integers, low and high (1 ≤ low ≤ high ≤ 109).

Output

Print n lines. The i-th of them should contain the result of the i-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.

Example
Input

题意：给你两个数，再给你一个范围，问是否在这个范围里存在这两个数的公约数。

思路：先用求出这两个数的最大公约数，然后求这个最大公约数的约数，再看范围内是否存在。

这里有一个快速求所有公约数的方法：

先求出 GCD(a , b ) = g；

然后找g的约数即可，（从1到sqrt（g）枚举）。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int cd[1000000];
int gcd(int a,int b){
    if(b==0)
        return a;
    else
        return(gcd(b,a%b));
}
int main()
{
    int a,b,n;
    scanf("%d%d",&a,&b);
    int maxn=gcd(a,b);
    int t=0;
    for(int i=1;i*i<=maxn;i++){
        if(maxn%i==0){
            cd[t++]=i;
            cd[t++]=maxn/i;
        }
    }
    sort(cd,cd+t);
    scanf("%d",&n);
    while(n--){
        int low,high,temp=0,ans=0;
        scanf("%d%d",&low,&high);
        for(int i=0;i<t;i++){
            if(cd[i]>=low&&cd[i]<=high){
                temp=1;
                ans=max(ans,cd[i]);
            }
        }
        if(temp)
            printf("%d\n",ans);
        else
            printf("-1\n");
        }
    return 0;
}