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;
}

<Sicily>Greatest Common Divisors的更多相关文章

  1. [UCSD白板题] Greatest Common Divisor

    Problem Introduction The greatest common divisor \(GCD(a, b)\) of two non-negative integers \(a\) an ...

  2. HDU 1423 Greatest Common Increasing Subsequence LCIS

    题目链接: 题目 Greatest Common Increasing Subsequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: ...

  3. POJ 2127 Greatest Common Increasing Subsequence -- 动态规划

    题目地址:http://poj.org/problem?id=2127 Description You are given two sequences of integer numbers. Writ ...

  4. HDOJ 1423 Greatest Common Increasing Subsequence -- 动态规划

    题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1423 Problem Description This is a problem from ZOJ 2 ...

  5. ZOJ 2432 Greatest Common Increasing Subsequence(最长公共上升子序列+路径打印)

    Greatest Common Increasing Subsequence 题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problem ...

  6. HDU 1423 Greatest Common Increasing Subsequence(最长公共上升LCIS)

    HDU 1423 Greatest Common Increasing Subsequence(最长公共上升LCIS) http://acm.hdu.edu.cn/showproblem.php?pi ...

  7. HDU1423:Greatest Common Increasing Subsequence(LICS)

    Problem Description This is a problem from ZOJ 2432.To make it easyer,you just need output the lengt ...

  8. greatest common divisor

    One efficient way to compute the GCD of two numbers is to use Euclid's algorithm, which states the f ...

  9. Greatest Common Increasing Subsequence hdu1423

    Greatest Common Increasing Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536 ...

随机推荐

  1. PHP中用下标符号[]去读取字符串的逻辑

    PHP中 [(下标)] 符号不仅能够应用于数组和对象,还能够应用于字符串,假设不注意非常easy出错. 比方获取一个网络接口,正常情况下会返回一个数组结构的json,经过解析之后结果为: array( ...

  2. python微框架Bottle(http)

    环境: win7系统 Python2.7 一 背景和概述 眼下项目中须要加入一个激活码功能,打算单独弄一个httpserver来写. 由于之前的游戏中已经有了一套完整的激活码生成工具和验证httpse ...

  3. HDU5638 / BestCoder Round #74 (div.1) 1003 Toposort 线段树+拓扑排序

    Toposort   问题描述 给出nn个点mm条边的有向无环图. 要求删掉恰好kk条边使得字典序最小的拓扑序列尽可能小. 输入描述 输入包含多组数据. 第一行有一个整数TT, 表示测试数据组数. 对 ...

  4. 基于FPGA的跨时钟域信号处理——专用握手信号

    在逻辑设计领域,只涉及单个时钟域的设计并不多.尤其对于一些复杂的应用,FPGA往往需要和多个时钟域的信号进行通信.异步时钟域所涉及的两个时钟之间可能存在相位差,也可能没有任何频率关系,即通常所说的不同 ...

  5. Linux下grub的配置文件

    GRUB(统一引导装入器)是基本的Linux引导装入器. 其有四个作用,如下: 1.选择操作系统(如果计算机上安装了多个操作系统). 2.表示相应引导文件所在的分区. 3.找到内核. 4.运行初始内存 ...

  6. hiho 172周 - 二维树状数组模板题

    题目链接 描述 You are given an N × N matrix. At the beginning every element is 0. Write a program supporti ...

  7. 贰、js的基础(二)类型转换

    JS 数据类型转换 方法主要有三种 转换函数.强制类型转换.利用js变量弱类型转换. 1. 转换函数: js提供了parseInt()和parseFloat()两个转换函数.前者把值转换成整数,后者把 ...

  8. 状压DP复习

    深感自己姿势水平之蒻……一直都不是很会状压DP,NOIP又特别喜欢考,就来复习一发…… 题目来源 Orz sqzmz T1 [BZOJ4197][NOI2015]寿司晚宴 (做过)质因数分解最大的质因 ...

  9. luogu P4062 [Code+#1]Yazid 的新生舞会(线段树+套路)

    今天原来是平安夜啊 感觉这题是道好题. 一个套路枚举权值\(x\),把权值等于\(x\)的设为1,不等于的设为-1,然后问题转化为多少个区间权值和大于. 发现并不是很好做,还有一个套路,用前缀和查分来 ...

  10. Couldn't connect to Docker daemon at http+docker://localunixsocket - is it running?

    解决方法: 1.进入启动文件目录 2.将用户加入到docker 组 sudo gpasswd -a ${USER}  docker 3.使用root用户 sudo su 4. 切换当前用户 su ${ ...