hdu 5974 A Simple Math Problem

最新推荐文章于 2020-11-19 20:12:12 发布

转载最新推荐文章于 2020-11-19 20:12:12 发布 · 172 阅读

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本文介绍了一种解决特定方程组的方法，即给定两个正整数a和b，寻找合适的X和Y使得X+Y等于a且X和Y的最小公倍数为b。文章提供了一个C++实现的示例程序，并详细解释了如何通过数学关系求解此问题。

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 data occupies one line,including two positive integers a(1≤a≤2*10^4),b(1≤b≤10^9),and their meanings are shown in the description.Contains most of the 12W test cases.

Output

For each set of input data,output a line of two integers,representing X, Y.If you cannot find such X and Y,output one line of "No Solution"(without quotation).

Sample Input

6 8
798 10780

Sample Output

No Solution
308 490


思路：
由题意可知：
                     lcm(x,y)=b;
因为             x*y =gcd(x,y)*lcm(x,y);
所以             x*y =gcd(x,y)*b;
因为             gcd(a,b)=gcd(x,y);
所以             x*y=b*gcd(a,b);
联立             x+y=a;

解个方程就行 ；



#include <cstdio>
#include <cstring>
#include <cmath>

int gcd (int m,int n){
	int flag;
	while(n>0){
		flag=m%n;
		m=n;
		n=flag;
	}
	return m;
}


int main (){
	int a,b;
	while(scanf("%d%d",&a,&b)!=EOF){
		int i,j;
		int c=gcd(a,b);
		int x,y;

		int delta= a*a-4*b*c;
		if(delta<0){
			printf("No Solution\n");
			continue;
		}
		int flag= sqrt(delta);
		if(flag*flag!=delta||(a-flag)%2!=0){
			printf("No Solution\n");
			continue;
		}

		x=(1*a-sqrt(delta))/2;
		y=a-x;

		printf("%d %d\n",x<y?x:y,x<y?y:x);

	}
	return 0;
}