codeforces 75C C. Modified GCD(二分）

题目链接:

codeforces 75C

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题目大意：

给出两个数，查询[l,r]内这两个数的gcd

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题目分析：

对这两个数的cd进行打表，然后二分查找即可。

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AC代码：

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <vector>using namespace std;int a,b,n,l,r;vector<int> v;int bsearch ( int x , int y ){    int l = 0 , r = v.size()-1,mid;    //cout << "YES : " << l << " " <<r << endl;    while ( l != r )    {        mid = (l+r+1)>>1;        //cout << v[mid] << " " << l << " " << r <<" " <<y << endl;        if ( v[mid] <= y ) l = mid;        else r = mid-1;    }    if ( v[l] >= x ) return v[l];    return -1;}int main ( ){    while (~scanf ("%d%d" , &a , &b ) )    {        for ( int i = 1 ; i*i <= a ; i++ )        {            if ( a%i ) continue;            if ( b%i == 0 ) v.push_back ( i );            int x = a/i;            if ( i == x ) continue;            if ( b%x == 0 ) v.push_back ( x );        }        sort ( v.begin() , v.end());        scanf ("%d" , &n );        while ( n-- )        {            scanf ( "%d%d" , &l , &r );            printf ( "%d\n" , bsearch ( l , r ));        }    }}