GCD and LCM HDU 4497 数论 题意 给你三个数x,y,z的最大公约数G和最小公倍数L,问你三个数字一共有几种可能.注意123和321算两种情况. 解题思路 L代表LCM,G代表GCD. \[ x=(p_1^{i_1})*(p_2^{i_2})*(p_3^{i_3})\dots \] \[ y=(p_1^{j_1})*(p_2^{j_2})*(p_3^{j_3})\dots \] \[ z=(p_1^{k_1})*(p_2^{k_2})*(p_3^{k_3})\dots \] \…

Problem Description Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and lcm(x, y, z) = L? Note, gcd(x, y, z) means the greatest common divisor of x, y and z, while lcm…

题目链接:https://vjudge.net/problem/HDU-4497 题意:求有多少组(x,y,z)满足gcd(x,y,z)=a,lcm(x,y,z)=b. 思路:对于x,y,z都可以写成x = p1^a1*p2^a2*p3^a3....pn^an;y = p1^b1*p2^b2*p3^b3....pn^bn;z=p1^c1*p2^c2*p3^c3....pn^cn;x,y,z的最大公约数可以写成gcd(x,y,z) = p1^min(a1,b1,c1)*p2^min(a2,b2,c…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4497 解题思路:将满足条件的一组x,z,y都除以G,得到x‘,y',z',满足条件gcd(x',y',x') = 1,同时lcm(x',y',x') = G/L. 特判,当G%L != 0 时,无解. 然后素数分解G/L,假设G/L = p1^t1 * p2^t2 *````* pn^tn. 满足上面条件的x,y,z一定为这样的形式. x' = p1^i1 * p2^i2 *```* pn^in.…

GCD and LCM Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2982    Accepted Submission(s): 1305 Problem Description Given two positive integers G and L, could you tell me how many solutions of…

GCD and LCM Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 3379    Accepted Submission(s): 1482 Problem Description Given two positive integers G and L, could you tell me how many solutions of…

GCD and LCM Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4497 Description Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and…

GCD and LCM Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 40    Accepted Submission(s): 22 Problem Description Given two positive integers G and L, could you tell me how many solutions of (x,…

GCD and LCM Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2024    Accepted Submission(s): 904 Problem Description Given two positive integers G and L, could you tell me how many solutions of (…

GCD and LCM Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2977    Accepted Submission(s): 1302 Problem Description Given two positive integers G and L, could you tell me how many solutions of…

并不重要的前言 最近学习了一些数论知识,但是自己都不懂自己到底学了些什么qwq,在这里把知识一并总结起来. 也不是很难的gcd和lcm 显而易见的结论: 为什么呢? 根据唯一分解定理: a和b都可被分解为素因子的乘积,形如: 则显而易见的有一下结论: 相乘,得: 得证 几种求gcd的算法 欧几里得算法(辗转相除法) 辗转相减法(优化:stein_gcd) 欧几里得算法 基于事实: 实现: int gcd(int a, int b){ ) ? a : gcd( b , a % b) ; } 简短而…

组合数学 GCD and LCM Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 451    Accepted Submission(s): 216 Problem Description Given two positive integers G and L, could you tell me how many solutions…

HDU4497 GCD and LCM 如果 \(G \% L != 0\) ,那么输出 \(0\) . 否则我们有 \(L/G=(p_1^{r_1})\cdot(p_2^{r_2})\cdot(p_3^{r_3})\cdots(p_m^{r_m})\) . 我们又有: \[ x=(p_1^{i_1})\cdot(p_2^{i_2})\cdot(p_3^{i_3})\cdots(p_m^{i_m}) \\ y=(p_1^{j_1})\cdot(p_2^{j_2})\cdot(p_3^{j_3})…

这个题目挺不错的,看到是通化邀请赛的题目,是一个很综合的数论题目. 是这样的,给你三个数的GCD和LCM,现在要你求出这三个数有多少种可能的情况. 对于是否存在这个问题,直接看 LCM%GCD是否为0,如果不为0的话,就没有满足条件的数哦,反之一定有. 接下来问题等价于求三个数GCD为1,LCM为LCM/GCD的种类数了. 设这个商为X. 首先我们可以把X因数分解成X=(p1*x1)*(p2*x2)*……*(pn*xn): 单独拿出一个素数进行讨论,如果要设ABC分别为满足情况的三个数,那么Xa…

题目链接: https://cn.vjudge.net/problem/23709/origin 本题其实有坑 数据大小太大, 2的32次方,故而一定是取巧的算法,暴力不可能过的 思路是最大公因数的倍数是最小公倍数,又有a <= b所以可以知道 a = gcd, b = lcm AC代码如下: #include <cstdio> #define ll long long using namespace std; int main() { int T; scanf("%d&quo…

GCD & LCM Inverse Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9928   Accepted: 1843 Description Given two positive integers a and b, we can easily calculate the greatest common divisor (GCD) and the least common multiple (LCM) of a a…

GCD and LCM Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 78 Accepted Submission(s): 43 Problem Description Given two positive integers G and L, could you tell me how many solutions of (x, y, z)…

题目链接: https://cn.vjudge.net/problem/POJ-2429 题目大意: 给出两个数的gcd和lcm,求原来的这两个数(限定两数之和最小). 解题思路: 首先,知道gcd和lcm求原来的两个数,需要分解lcm / gcd .将其分解为互质的两个数. 首先将lcm/gcd质因数分解,要分解出沪互质两个数字,那么这两个数字的gcd=1,也就是没有公共的质因子,所以可以直接枚举这两个数字的质因子,如果一个数要取这个质因子,就把它的指数全部取掉. 质因数分解用大数因式分解来做…

GCD and LCM Descriptions: Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b. Input Input consists of several data sets. Each data set contains a and b separated by a single space in…

WUSTOJ 1266: gcd和lcm 参考 1naive1的博客 Description   已知a,b的最大公约数为x,也即gcd(a,b)=x; a,b的最小公倍数为y,也即lcm(a,b)=y.给出x,y.求满足要求的a和b一共有多少种. Input   多组测试样例.每组给两个整数x,y.(1<=x<=100000,1<=y<=1000000000). Output   对于每个测试样例,输出一个整数,表示满足要求的(a,b)的种数. Sample Input 3 60…

GCD _ LCM 是给你两个数A B 的最大公约数, 以及最小公倍数 the greatest common divisor and the least common multiply ! 最大公约数最简单.最常见的算法,就是辗转相除法   : 假设 GCD(A , B) ; A / B = P ;  A % B = Q;  那么 A =  B P + Q; GCD(B, Q); GCD (A , B)  % GCD (B , Q) = 0     :  因为 A  的表达式当中包括了 B .…

Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and lcm(x, y, z) = L? Note, gcd(x, y, z) means the greatest common divisor of x, y and z, while lcm(x, y, z) means the…

题意:三个数x, y, z. 给出最大公倍数g和最小公约数l.求满足条件的x,y,z有多少组. 题解:设n=g/l n=p1^n1*p2^n2...pn^nk (分解质因数 那么x = p1^x1 * p2^x2 * .... ^ pn^xk y = p1^y1 * p2^y2 * .... ^ pn^yk x = p1^z1 * p2^z2 * .... ^ pn^zk 那么对于任意i (0<=i<=k) 都有 min(xi, yi, zi) = 0, max(xi, yi, zi) = n…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4497 题意:已知GCD(x, y, z) = G,LCM(x, y, z) = L.告诉你G.L,求满足要求的(x, y, z)有多少组,并且要考虑顺序. 思路:如果L%G != 0显然不存在这样的(x, y, z),相反肯定存在.具体做法就是将L/G分解质因子,得到:L/G = P1^t1 * P2^t2 * ... * Pk^tk,我们来考虑任意一个因子Pi^ti,此时(x/G, y/G, z/…

质分解 + 简单计数.当时去比赛的时候太年轻了...这道题都没敢想.现在回过头来做了一下,发现挺简单的,当时没做这道题真是挺遗憾的.这道题就是把lcm  / gcd 质分解,统计每个质因子的个数,然后就可以统计出总数了. 统计的时候假如有2个3,这样的话肯定是有一个元素是含有全部的2个3的,也肯定有一个元素没有3,于是我们就可以直接得出,统计个数为元素个数x6, 然后每个质因子分配情况互不影响,于是可以用乘法原理.就可以得出最终答案了. #include<algorithm> #include…

链接 :  http://acm.hdu.edu.cn/showproblem.php?pid=4497 假设G不是L的约数 就不可能找到三个数. L的全部素因子一定包括G的全部素因子 而且次方数一定大于等于G的.仅仅须要三个数 对于每个素因子的次方数 三个的最小值是G的,最大值是L的.考虑三个相应的次方数都不一样.那么当中两个是确定的 一个是G的一个是L的 剩下的一个在G和L的之间. 算上排列 总共同拥有6种.或者当中两个是一样的,那么也有6种情况. 最后能够合并计算. //#pragma…

A frog has just learned some number theory, and can't wait to show his ability to his girlfriend. Now the frog is sitting on a grid map of infinite rows and columns. Rows are numbered 1,2,⋯from the bottom, so are the columns. At first the frog is sit…

题意: 给两个数,lll 和 ggg,为x , y , z,的最小公倍数和最大公约数,求出x , y , z 的值有多少种可能性 思路: 将x , y , z进行素因子分解 素因子的幂次 x a1 a2 a3 a4 y b1 b2 b3 b4 z c1 c2 c3 c4 gcd min(a1,b1,c1) min(a2,b2,c3)- lcm max(a1,b1,c1) max(a2,b2,c3)- 第一组样例: 6=21 * 31 72= 23 * 32 最大公约数和最小公倍数约分得 12=2…

Describtion First we define: (1) lcm(a,b), the least common multiple of two integers a and b, is the smallest positive integer that is divisible by both a and b. for example, lcm(2,3)=6 and lcm(4,6)=12. (2) gcd(a,b), the greatest common divisor of tw…

题意:给定G,L,分别是三个数最大公因数和最小公倍数,问你能找出多少对. 析:数学题,当时就想错了,就没找出规律,思路是这样的. 首先G和L有公因数,就是G,所以就可以用L除以G,然后只要找从1-(n=L/G),即可,那么可以进行质因数分解,假设: n = p1^t1*p2^t2*p3^t3;那么x, y, z,除以G后一定是这样的. x = p1^i1*p2^i2*p3^i3; y = p1^j1*p2^j2*p3^j3; z = p1^k1*p2^k2*p3^k3; 那么我们可以知道,i1,…