HDU 5974 A Simple Math Problem (解方程) 目录作者:@dwtfukgv本文为作者原创,转载请注明出处:https://www.cnblogs.com/dwtfukgv/p/6793045.html题意:给定a和b,求一组满足x+y=a && lcm(x, y)=b。析:x+y = a, lcm(x, y) = b,=>x + y = a, x * y = b * k,其中 k = gcd(x, ...
hdu5974 A Simple Math Problem(数学) 题目链接 大意:给你两个数X,YX,Y,让你找两个数a,ba,b,满足a+b=X,lcm(a,b)=Ya+b=X,lcm(a,b)=Y. 思路:枚举gcd(a,b)gcd(a,b),假设gcd(a,b)=k,那么a=xa∗k,b=xb∗k,gcd(a,b)=k,那么a=xa∗k,b=xb∗k,化简上面给的两个式子...
A Simple Math Problem HDU - 5974(数论,gcd相关的公式推导,好题),题意:A=X+Y且X与Y的最小公倍数是B,已知A、B,求X、Y。题解:参考博客先说几个才知道的性质性质1:如果k1,k2互质,则k1+k2与k1*k2也是互质的(具体证明可以点击上面的博客,用的是反正法)。对于本题其
那么容易得到:x*y等于b*gcd(a,b) 此时,x,y的和与积已知,联立求解,最后判断是否可以得到整数解 ps: 一条性质:如果x与y互质,那么x*y和x+y肯定互质 #include<stdio.h> #include<math.h> long long gcd(long long a,long long b) { long long r; while(b) { r=a%b; a=b; b=r; } return ...
HDU 5974 A Simple Math Problem 数学题 http://acm.hdu.edu.cn/showproblem.php?pid=5974 遇到数学题真的跪。。 题目要求 X + Y = a lcm(X, Y) = b 设c = gcd(x, y); 那么可以表达出x和y了,就是x = i * c; y = j * c;...
题目简介 Given two positive integers a and b,find suitable X and Y to meet the conditions:X...
Problem Description Given two positive integers a and b,find suitable X and Y to meet the conditions: X+Y=a Least Common Multiple (X, Y) =b Input Input includes multiple sets of test data.Each test da...HDU 5974 A Simple Math Problem Problem Description Given two positive integers a ...
HDU - 1757 A Simple Math Problem(矩阵快速幂,水题) 题目链接:点击查看 题目大意:实现公式: f(x)=x,x<10 f(x)=a0*f(x-1)+a1*f(x-2)+……+a9*f(x-10) 题目给出a0~a9,一个n和一个m,要求输出f(n)对m取模后的结果 题目分析:水题 初始矩阵:(取n=10即可) f(n-1) f(n-2) f(n...
A Simple Math Problem HDU - 5974 转自http://www.cnblogs.com/kimsimple/p/6792395.html 题意 Given two positive integers a and b,find suitable X and Y to meet the conditions: X+Y=a Least Common Multiple (X, Y) =b 分析 x*y = b*gcd(x,y); ...
#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>usingnamespacestd; typedeflonglongll; ll gcd(ll a,ll b) {returnb==0?a:gcd(b,a%b); }intmain() { ll a,b;while(scanf("%lld%lld",&a,&b)!=EOF) ...