CodeForces 75C Modified GCD

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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 andb 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 integern, 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.

Examples
input
9 2731 510 119 11
output
3-19

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。

刚开始编译器出点故障一直没输出。。害的我半天不知道哪出问题了。无语。。。。。。

欧几里德算法的几种代码实现

递归法：

int gcd(int a,int b) {    if(b==0)         return a;     return          gcd(b,a%b); }

优化后

int gcd(int a,int b){    return b ? gcd(b,a%b) : a;

迭代形式

int Gcd(int a, int b)  {     while(b != 0)    {    　　int r = b;      　　b = a % b;      　　a = r;    }      return a; }

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;int gcd(int a,int b){int m;while(a){m=b%a;b=a;a=m;    }    return b; } int main(){int a,b,n,i,j,l,k,num,low,high,ab[1100];while(scanf("%d%d",&a,&b)!=EOF){int ans=gcd(a,b);k=0;    for(i=1;i*i<=ans;i++)    {   if(ans%i==0)   {  ab[k++]=i;  if(i*i!=ans)  { ab[k++]=ans/i;  }   }    }    sort(ab,ab+k);        scanf("%d",&n);     while(n--)    {   int flag=0;   scanf("%d%d",&low,&high);   for(i=k-1;i>=0;i--)   {           if(ab[i]<=high&&ab[i]>=low)      {     printf("%d\n",ab[i]);     flag=1;     break;   }   }   if(flag==0)   printf("-1\n");   }}return 0;}

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