hdu5974 A Simple Math Problem(数学)

题目链接

大意：给你两个数X,YX,YX,Y,让你找两个数a,ba,ba,b，满足a+b=X,lcm(a,b)=Ya+b=X,lcm(a,b)=Ya+b=X,lcm(a,b)=Y.

思路：枚举gcd(a,b)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=x_a*k,b=x_b*k,gcd(a,b)=k，那么a=xa​∗k,b=xb​∗k,化简上面给的两个式子即可得到

xa+xb=Xk,xa∗xb=Ykx_a+x_b=\frac{X}{k},x_a*x_b=\frac{Y}{k}xa​+xb​=kX​,xa​∗xb​=kY​，显然这是一个方程组，将xb=Xk−xax_b=\frac{X}{k}-x_axb​=kX​−xa​代入第二个式子即可得到一个一元二次方程，解这个方程即可。

#include<bits/stdc++.h>

#define LL long long
#define fi first
#define se second
#define mp make_pair
#define pb push_back

using namespace std;

LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
const int N = 2e5 +11;
int a,b;
int main(){
	ios::sync_with_stdio(false);
	while(cin>>a>>b){
		int q=sqrt(a);
		int sta=0,anL,anR;
		for(int i=1;i<=q;i++){
			if(a%i)continue;
			int A=a/i;
			int B=b/i;
			if(A*A<4*B)continue;
			int T=floor(sqrt(A*A-4*B));
			if(T*T!=A*A-4*B)continue;
			else {
				if((A-T)%2||(A+T)%2)continue;
				int L=(A-T)/2*i;
				int R=(A+T)/2*i;
				if(lcm(L,R)!=b)continue;
				anL=(A-T)/2*i;
				anR=(A+T)/2*i;
				sta=1;
				break;
			}
		}
		for(int i=1;i<=q&&!sta;i++){
			if(a%i)continue;
			int A=a/(a/i);
			int B=b/(a/i);
			if(A*A<4*B)continue;
			int T=floor(sqrt(A*A-4*B));
			if(T*T!=A*A-4*B){continue;}
			else {
				if((A-T)%2||(A+T)%2){continue;}
				int L=(A-T)/2*(a/i);
				int R=(A+T)/2*(a/i);
				if(lcm(L,R)!=b){continue;}
				anL=(A-T)/2*(a/i);
				anR=(A+T)/2*(a/i);
				sta=1;
				break;
			}
		}
		if(!sta)cout<<"No Solution\n";
		else cout<<anL<<' '<<anR<<endl;
	}
	return 0;
}