欧几里德和扩展（hdu 1222）

最新推荐文章于 2018-09-05 17:41:24 发布

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本文介绍了使用欧几里德算法求解最大公约数的方法，包括基本的欧几里德算法和扩展欧几里德算法，并通过实例展示了如何应用于解决实际问题。

欧几里德：求两个数的最大公约数。gcd(a,b) = gcd(b,a mod b)

代码：

int gcd(int x,int y){
	return x%y==0?y:gcd(y,x%y);

}

扩展欧几里德：对于不完全为 0 的非负整数 a，b，gcd（a，b）表示 a，b 的最大公约数，必然存在整数对 x，y ，使得 gcd（a，b）=ax+by。

用途：1.求不定方程：ax+by=c. 2.求解模线性方程 ( 线性同余方程 ) 3.求解模的逆元。
基本方法：递归

思路：c %(a,b)=0,则该方程有解。由 gcd(a,b)=gcd(b,a%b),

则gcd = ax1+by1 = bx2+(a%b)y2 = bx2+(a-(a/b)*b)*y2 = ay2+b(x-a/b*y2).

等式求解为 x1 = y2 , y1 = x2-a/b*y2 . 递归结束条件为b==0,此时x=1.

代码：

int exgcd(int a,int b,int &x,int &y){
	if (b==0){
		x=1;y=0;
		return a;
	}
	int ans = exgcd(b,a%b,x,y);
	int temp = x;
	x = y;
	y = temp-a/b*y;
	return ans;	
}//x,y为一组特解x0,y0.   x = x0+(b/gcd)*i,y = y-(a/gcd)*i.

求解不定方程，代码：

bool jie(int a,int b,int c,int &x,int &y){
	int d = exgcd(a,b,x,y);
	if (c%d!=0) return false;
	int k=c/d;
	x*=k;
	y*=k;
	return true;
}

简单例题：hdu 1222

原题

Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7377 Accepted Submission(s): 3700

Problem Description

There is a hill with n holes around. The holes are signed from 0 to n-1.

A rabbit must hide in one of the holes. A wolf searches the rabbit in anticlockwise order. The first hole he get into is the one signed with 0. Then he will get into the hole every m holes. For example, m=2 and n=6, the wolf will get into the holes which are signed 0,2,4,0. If the rabbit hides in the hole which signed 1,3 or 5, she will survive. So we call these holes the safe holes.

Input

The input starts with a positive integer P which indicates the number of test cases. Then on the following P lines,each line consists 2 positive integer m and n(0<m,n<2147483648).

Output

For each input m n, if safe holes exist, you should output "YES", else output "NO" in a single line.

Sample Input

Sample Output

  
   NO
YES

题意：狼从0开始，每隔m个洞进洞，共n个洞，问有没有安全洞，即狼不会进的洞。

题解：找规律，若m，n的最大公约数大于1(即m,n不互质)，就有安全洞。

代码：

#include<cstdio>
using namespace std;

int gcd(int a,int b){
	if (a%b == 0)
	return b;
	return gcd(b,a%b);
}
int main(){
	int t,n,m;
	scanf("%d",&t);
	while (t--){
		scanf("%d %d",&m,&n);
		if (gcd(m,n)>1)
		printf ("YES\n");
		else
		printf ("NO\n");
	}
	return 0;
}