Blocks to Cubes
Bholu the Pandit on this New Year wanted to divide his Cuboidal Packaging block into cubes. But he loves uniformity so he asks you to divide it such a way that all the cubes are of same size and volume of individual cube is as large as possible.
Note: He will utilize whole volume i.e volume of cuboid before dividing is same as sum of volume of all the cubes.
Input
The first input line contains an integer T, the number of testcases. Each testcase consist of single line which consist of 3 space separated integers a, b & c representing length, breadth and height of the Cuboidal block.
Output
For each testcase you need to output 2 space separated integers, the length of side of the cube and the number of cubes which could be formed out of this cuboidal packaging block. As the number of cubes which could be made could be a large number so just output the answer modulus 109+7 (1000000007).
Constraints
1 ≤ T ≤ 1000
1 ≤ a,b,c ≤ 109
2
2 4 6
1 2 3
2 6
1 6
In the 1st testcase a=2, b=4 & c=6. So length of the side of the cube would be 2 and the number of cubes which would be formed would be 6. In the 2nd testcase a=1, b=2 & c=3. So length of the side of the cube would be 1 and the number of cubes which would be formed would be 6.
Approach # 1:
/*
// Sample code to perform I/O: #include <iostream> using namespace std; int main() {
int num;
cin >> num; // Reading input from STDIN
cout << "Input number is " << num << endl; // Writing output to STDOUT
} // Warning: Printing unwanted or ill-formatted data to output will cause the test cases to fail
*/ // Write your code here
#include<iostream>
#include<algorithm>
const int mod = 1e9 + 7; using namespace std; long long gcd(long long x, long long y) {
if (y == 0)
return x;
else return gcd(y, x%y);
} int main() {
int n;
long long l, w, h;
long long ans;
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> l >> w >> h;
long long c = gcd(gcd(l, w), h);
l /= c;
w /= c;
h /= c;
ans = (((l * w) % mod) * h) % mod;
cout << c << ' ' << ans << endl;
} return 0;
}
Analysis:
From this problem I learned how to deal with the three numbers problem. And we shuold use long long.
Blocks to Cubes的更多相关文章
- Intel® Threading Building Blocks (Intel® TBB) Developer Guide 中文 Parallelizing Data Flow and Dependence Graphs并行化data flow和依赖图
https://www.threadingbuildingblocks.org/docs/help/index.htm Parallelizing Data Flow and Dependency G ...
- POJ 1052 Plato's Blocks
Plato's Blocks Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 734 Accepted: 296 De ...
- Codeforces Round #356 (Div. 2) D. Bear and Tower of Cubes dfs
D. Bear and Tower of Cubes 题目连接: http://www.codeforces.com/contest/680/problem/D Description Limak i ...
- codeforces 680D D. Bear and Tower of Cubes(dfs+贪心)
题目链接: D. Bear and Tower of Cubes time limit per test 2 seconds memory limit per test 256 megabytes i ...
- Codeforces Round #295 D. Cubes [贪心 set map]
传送门 D. Cubes time limit per test 3 seconds memory limit per test 256 megabytes input standard input ...
- 从Script到Code Blocks、Code Behind到MVC、MVP、MVVM
刚过去的周五(3-14)例行地主持了技术会议,主题正好是<UI层的设计模式——从Script.Code Behind到MVC.MVP.MVVM>,是前一天晚上才定的,中午花了半小时准备了下 ...
- 【POJ-1390】Blocks 区间DP
Blocks Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 5252 Accepted: 2165 Descriptio ...
- 开发该选择Blocks还是Delegates
前文:网络上找了很多关于delegation和block的使用场景,发现没有很满意的解释,后来无意中在stablekernel找到了这篇文章,文中作者不仅仅是给出了解决方案,更值得我们深思的是作者独特 ...
- 水泡动画模拟(Marching Cubes)
Marching Cubes算法是三维离散数据场中提取等值面的经典算法,其主要应用于医学领域的可视化场景,例如CT扫描和MRI扫描的3D重建等. 算法主要的思想是在三维离散数据场中通过线性插值来逼近等 ...
随机推荐
- url_encode and url_decode in Shell
之前写过一版 shell下解码url,下面给出另外一个版本 from https://gist.github.com/cdown/1163649 function urlencode() { loca ...
- 13-Oulipo(kmp裸题)
http://acm.hdu.edu.cn/showproblem.php?pid=1686 Oulipo Time Limit: 3000/1000 MS (Java/Others) Memo ...
- freemaker 优缺点 及 应用配置
通俗的讲,freemaker其实就是一个模板引擎.什么意思呢?——Java可以基于依赖库,然后在模板上进行数据更改(显示). 在模板中,您专注于如何呈现数据,而在模板外(后台业务代码),您将专注于呈现 ...
- WebAPI的路由规则
1.自定义路由 public static class WebApiConfig { public static void Register(HttpConfiguration config) { / ...
- Openssl base64命令
一.简介 对文件件进行base64的编码与解码 二.语法 openssl base64 [-in filename] [-out filename] 三.实例 1.二进制文件与base64编码互转 o ...
- STL中mem_fun, mem_fun_ref用法
1.引言 先看一个STL中for_each的用法: #include <iostream> #include <vector> #include <algorithm&g ...
- [C++] Memory Retrieval(内存检索)
Traverse the memory by (char*) , because every time it will increase by 1byte when i want get the i ...
- python2.7 跨文件全局变量的方法-乾颐堂
在使用Python编写的应用的过程中,有时会遇到多个文件之间传递同一个全局变量的情况. 文件1:globalvar.py 1 2 3 4 5 6 7 8 9 10 11 12 #!/usr/bin/e ...
- 访问SAP的RFC
.NET 环境Xp(sp3) vs2010, win2003 EN 32bit(sp2)winform,webform 引用sapnco.dll,sapnco_utils.dll(自动引用)配置文件需 ...
- QT之Variant
QVariant识别类型的注册 QVariant识别类型的注册 QVariant为一个万能的数据类型--可以作为许多类型互相之间进行自动转换.将C++变为弱数据类型成为可能--也是许多控件中用户定义数 ...