UVA11388 GCD LCM(数论)

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UVA11388 GCD LCM(数论)

题目链接. 题意: 给定两个数,一个G,一个L,找出两个数a,b(a<=b),使得这两个数的最大公约数为G,最小公倍数为L,且(a最小). 分析: 当a,b存在时,a一定为G. 自己证了一下,数学方面不太擅长. 假设 a 最小为 k1G (其中 k1 != 1), b为 k2G, 即 a = G,不满足条件. 那么a*b=k1*k2*G^2=L*G 这时一定有 a1 = G, b2 = k1*k2G 满足条件.即a不符合题意. 设 a = G, b = kG 因为 L*G = a*b b = L…

洛谷 UVA11388 GCD LCM

UVA11388 GCD LCM Description of the title PDF The GCD of two positive integers is the largest integer that divides both the integers without any remainder. The LCM of two positive integers is the smallest positive integer that is divisible by both th…

UVA11388 GCD LCM

(链接点这儿) 题目: The GCD of two positive integers is the largest integer that divides both the integers without any remainder. The LCM of two positive integers is the smallest positive integer that is divisible by both the integers. A positive integer can…

2015多校第8场 HDU 5382 GCD?LCM! 数论公式推导

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5382 题意:函数lcm(a,b):求两整数a,b的最小公倍数:函数gcd(a,b):求两整数a,b的最大公约数.函数[exp],其中exp是一个逻辑表达式.如果逻辑表达式exp是真,那么函数[exp]的值是1,否则函数[exp]的值是0.例如:[1+2>=3] = 1 ,[1+2>=4] = 0. 求S(n)的值. #include <bits/stdc++.h> using name…

数论入门2——gcd,lcm,exGCD,欧拉定理,乘法逆元,(ex)CRT,(ex)BSGS,(ex)Lucas,原根,Miller-Rabin,Pollard-Rho

数论入门2 另一种类型的数论... GCD,LCM 定义\(gcd(a,b)\)为a和b的最大公约数,\(lcm(a,b)\)为a和b的最小公倍数,则有: 将a和b分解质因数为\(a=p1^{a1}p2^{a2}p3^{a3}...pn^{an},b=p1^{b1}p2^{b2}p3^{b3}...pn^{bn}\),那么\(gcd(a,b)=\prod_{i=1}^{n}pi^{min(ai,bi)},lcm(a,b)=\prod_{i=1}^{n}pi^{max(ai,bi)}\)(0和任何…