<Sicily>Greatest Common Divisors
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
- [UCSD白板题] Greatest Common Divisor
Problem Introduction The greatest common divisor \(GCD(a, b)\) of two non-negative integers \(a\) an ...
- 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: ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- greatest common divisor
One efficient way to compute the GCD of two numbers is to use Euclid's algorithm, which states the f ...
- Greatest Common Increasing Subsequence hdu1423
Greatest Common Increasing Subsequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536 ...
随机推荐
- "浪潮杯"第六届ACM山东省省赛山科场总结
从空间拷过来的.尽管已经过去一个月了.记忆犹新 也算是又一次拾起这个blog Just begin 看着一群群大牛还有队友男神的省赛总结都出了 我最终也耐不住寂寞 来做个流水账抒抒情好了 第一次省赛 ...
- 关于 折半查找 while 条件 < , <=
int bin_search(int a[],int len,int key) { int low=0; int high=len-1; while(low<=high) //若为low< ...
- netty底层是事件驱动的异步库 但是可以await或者sync(本质是future超时机制)同步返回 但是官方 Prefer addListener(GenericFutureListener) to await()
io.netty.channel 摘自:https://netty.io/4.0/api/io/netty/channel/ChannelFuture.html Interface ChannelFu ...
- Win7 如何禁用“切换用户”功能
1.按win+r,输入gpedit.msc,点击确定: 2.依次点击计算机配置--管理模块--系统--登录,右侧列表中找到“隐藏“快速用户切换”的入口点”: 3.双击隐藏“快速用户切换”的入口点,点击 ...
- 在网页html中嵌入特殊字体
1.字体格式 .EOT,适用于Internet Explorer 4.0+ .TTF或.OTF,适用于Firefox 3.5.Safari.Opera .SVG,适用于Chrome.IPhone 最常 ...
- POJ3764 The xor-longest Path(字典树)
题意 给你一棵树,n个节点,n-1条边每条边i都有一个权值wi.定义任意两点间的权值为:这两点间的路径上的所有边的值的异或.比如a点和b点间有i,j,k三条边,那么ab两点间的权值为:wi^wj^wk ...
- jQuery第三课 点击按钮 弹出层div效果
jQuery 事件方法 事件方法会触发匹配元素的事件,或将函数绑定到所有匹配元素的某个事件. 触发实例: $("button#demo").click() 上面的例子将触发 id= ...
- VUEJS开发规范
VUEJS开发规范 基于组件化开发理解 组件命名规范 结构化规范 注释规范 编码规范 基于组件化开发理解 什么是组件? ``` 组件其实就是页面组成的一部分,好比是电脑中的每一个元件(如硬盘.键盘.鼠 ...
- 紫书 习题 8-2 UVa 1610 (暴力出奇迹)
这道题我真的想的非常的复杂, 拿草稿纸一直在找规律,推公式, 然后总有一些特殊的情况. 然后就WA了N次.无奈之下看了别人的博客, 然后就惊了.直接暴力枚举两个相邻字符串 里面的所有可能就可以了--真 ...
- HDU 4069 数独
好久没做题了,建图搞了好久…… 然后,判是否有多解的时候会把原来的答案覆盖掉…… 这里没注意,弄了一下午…… 代码: #include <iostream> #include <cs ...