<Sicily>Greatest Common Divisors

最新推荐文章于 2021-02-19 12:48:11 发布

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#sicily #GCD

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该博客介绍了一道关于寻找两个正整数a和b在指定范围low到high内最大公约数的问题。如果范围内没有符合条件的公约数，则输出"No answer"。文章提供了问题描述、输入输出格式，并给出了解题思路和代码实现。

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一、题目描述

A common divisor for two positive numbers is a number which both numbers are divisible by. It’s easy to calculate the greatest common divisor between tow numbers. 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.

二、输入

The first line contains an integer T (1<=T<=10)- indicating the number of test cases.
For each case, there are four integers a, b, low, high (1<=a,b<=1000,1<=low<=high<=1000) in one line.

三、输出

For each case, print the greatest common divisor between a and b in given range, if there is no common divisor in given range, you should print “No answer”(without quotes).
Sample Input

四、解题思路

题意：从low到high之间找出既能被a整除，又能被b整除的数，如果没有输出No answer

思路：这道题没什么好讲，就是遍历从high到low开始找一个既能被a整除又能被b整除就行了。

五、代码

#include<iostream>

using namespace std;

int main()
{
    int times;
    cin >> times;
    while(times--)
    {
        int a, b, low, high;

        cin >> a >> b >> low >> high;

        bool result;
        int divisor;

        for(divisor = high; divisor >= low; divisor--)
        {
            if(a % divisor == 0 && b % divisor == 0) {result = true; break;}
            result = false;
        }

        if(result) cout << divisor << endl;
        else cout << "No answer" << endl;
    }
    return 0;
}