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
9 27
3
1 5
10 11
9 11
3
-1
9
题解：找出a,b的所有公共约数，然后二分查找区间左右端点的位置，分类讨论
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int maxn=2e5+10; 5 6 int a,b,n; 7 int ma[maxn]; 8 9 int gcd(int a,int b){ 10 if(b==0) return a; 11 return gcd(b,a%b); 12 } 13 14 int main() 15 { cin>>a>>b>>n; 16 int g=gcd(a,b),cnt=0; 17 for(int j=1;j<=sqrt(g)+1;j++){ 18 if(g%j!=0) continue; 19 ma[cnt++]=j; 20 if(j<g/j) ma[cnt++]=g/j; 21 } 22 sort(ma,ma+cnt); 23 int aa,bb; 24 for(int i=1;i<=n;i++){ 25 cin>>aa>>bb; 26 //for(int j=0;j<cnt;j++) cout<<ma[j]<<endl; 27 int pos1=lower_bound(ma,ma+cnt,aa)-ma; 28 int pos2=lower_bound(ma,ma+cnt,bb)-ma; 29 //cout<<pos1<<" "<<pos2<<endl; 30 if(pos1==pos2){ 31 if(ma[pos2]==bb) cout<<bb<<endl; 32 else cout<<"-1"<<endl; 33 } 34 else{ 35 if(ma[pos2]==bb) cout<<ma[pos2]<<endl; 36 else cout<<ma[pos2-1]<<endl; 37 } 38 } 39 return 0; 40 }