题目描述

Description

Having decided to invest in renewable energy, Byteasar started a solar panels factory. It appears that he has hit the gold as within a few days clients walked through his door. Each client has ordered a single rectangular panel with specified width and height ranges.
The panels consist of square photovoltaic cells. The cells are available in all integer sizes, i.e., with the side length integer, but all cells in one panel have to be of the same size. The production process exhibits economies of scale in that the larger the cells that form it, the more efficient the panel. Thus, for each of the ordered panels, Byteasar would like to know the maximum side length of the cells it can be made of.
n组询问，每次问smin<=x<=smax, wmin<=y<=wmax时gcd(x, y)的最大值。

Input

The first line of the standard input contains a single integer N(1<=N<=1000): the number of panels that were ordered. The following lines describe each of those panels: the i-th line contains four integers Smin,Smax,Wmin,Wmax(1<=Smin<=Smax<=10^9,1<=Wmin<=Wmax<=10^9), separated by single spaces; these specify the minimum width, the maximum width, the minimum height, and the maximum height of the i-th panel respectively.

Output

Your program should print exactly n lines to the standard output. The i-th line is to give the maximum side length of the cells that the i-th panel can be made of.

Sample Input

4
3 9 8 8
1 10 11 15
4 7 22 23
2 5 19 24

Sample Output

8
7
2
5

HINT

Explanation: Byteasar will produce four solar panels of the following sizes: 8*8 (a single cell), 7*14 (two cells), 4*22 or 6*22 (22 or 33 cells respectively), and 5*20 (four cells).

题目分析

结论:出现的倍数的条件
然后分块枚举 判断同时在两段区间里出现即可

#include <cstdio>
#include <cstring>
#include <set>
#include <map>
#include <vector>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
int t;
int n,m;
int main()
{   
    scanf("%d",&t);
    while(t--)
    {
        int a,b,c,d;
        scanf("%d%d%d%d",&a,&b,&c,&d);
        a--,c--;
        int last=0,ans=0;
        for(int i=1;i<=b&&i<=d;i=last+1)
        {
            last=min(b/(b/i),d/(d/i));
            if(b/last>a/last&&d/last>c/last) ans=last;
        }
        printf("%d\n",ans);
    }
    return 0;
}