acd LCM Challenge(求1~n的随意三个数的最大公倍数)

【acd LCM Challenge(求1~n的随意三个数的最大公倍数)】的更多相关文章

acd LCM Challenge(求1~n的随意三个数的最大公倍数)

Problem Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it. But I also don't want to use many numbers, so I'll choose three positive integers…

tr循环，每行 2个数相加求出和位第三个数赋值 (http://jsfiddle.net/hgeL44rz/113/)

<table id="tb"> <tr> <th>单价</th> <th>数量</th> <th>小计</th> </tr> <tr> <td><input type="text" value="2" class="input" readonly style="border:non…

A - LCM Challenge

A - LCM Challenge Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) Submit Status Problem Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I w…

LightOj 1215 - Finding LCM（求LCM(x, y)=L中的 y ）

题目链接:http://lightoj.com/volume_showproblem.php?problem=1215 题意:已知三个数a b c 的最小公倍数是 L ,现在告诉你 a b L 求最小的 c ; 其实就是告诉你(最小公倍数LCM)LCM(x, y) = L 已知 x 和 L 求最小的 y ; L = LCM(x, y)=x*y/gcd(x, y);如果把x,y,L写成素因子之积的方式会很容易发现 L 就是 x 和 y 中素因子指数较大的那些数之积; 例如LCM(24, y)…

acdream.LCM Challenge(数学推导）

LCM Challenge Time Limit:1000MS Memory Limit:64000KB 64bit IO Format:%lld & %llu Submit Status Practice ACdream 1077 Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I…

[codeforces 235]A. LCM Challenge

[codeforces 235]A. LCM Challenge 试题描述 Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it. But I also don't want to use many numbers, so I'll choose three…

Codeforces Round #146 (Div. 1) A. LCM Challenge 水题

A. LCM Challenge 题目连接: http://www.codeforces.com/contest/235/problem/A Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it. But I also don't wa…