GCD(一) 题目: The greatest common divisor GCD(a,b) of two positive integers a and b,sometimes written (a,b),is the largest divisor common to a and b,For example,(1,2)=1,(12,18)=6. (a,b) can be easily found by the Euclidean algorithm. Now Carp is conside
gcd算法是用来求两个数最大公约数的算法,他是依靠辗转相除(中国好像叫辗转相减)法来求两个数的最大公约数,别的地方也有很多介绍不做过多赘述,主要提供代码供自己参考. gcd(int a,int b) { ?a:gcd(b,a%b); } 对于拓展gcd,是对方程ax+by=c求解: 就是a mod b=c这个方程求解: 具体看代码,不做过多赘述因为别人已经讲的很详细了,在这里copy一下别人的讲解: 我们其实只需要考虑形如 a • x mod n = 1 的方程.因为,如果能解出这样的方程, a
Rastas's has been given a number n. Being weak at mathematics, she has to consider all the numbers from 1 to 2n - 1 so as to become perfect in calculations. (You can assume each number is consider as a soldier). We define the strength of number i as
gcd(a, b)用于求解自然数a,b的最大公约数 int gcd(int a, int b) { ) return a; return gcd(b, a%b); } extgcd(a, b, x, y)用于求解方程ax+by = 1的一组解,并返回a,b的最大公约数 int exgcd(int a, int b, int &x, int &y) { int d = a; ) { d = exgcd(b, a%b, y, x); y -= (a/b)*x; } else { x = ; y
题目链接 http://acm.split.hdu.edu.cn/showproblem.php?pid=5869 Problem Description This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting.
