A Simple Math Problem 题解(数论)

题目链接

题目大意

给你\(a(1<=a<=2e4),b\)(1<=b<=1e9)

要你找到x和y使得\(x+y=a,lcm(x,y)=b\)

题目思路

这个主要是要明白

\(gcd(x,y)=gcd(a,b)\)

设\(x=nk,y=mk\)

则\(gcd(a,b)=gcd((n+m)k,nmk)\)

而\(gcd(n,m)=1\)推到出\(gcd(n+m,nm)=1\)

\(xy=lcm(x,y)*gcd(x,y)=b*gcd(a,b)\)

而已知\(x+y=a\)

则可以推出x和y

代码

#include<bits/stdc++.h>
#define fi first
#define se second
#define debug cout<<"I AM HERE"<<endl;
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn=1e2+5,inf=0x3f3f3f3f,mod=1e9+7;
const int eps=1e-3;
int a,b;
signed main(){
    while(scanf("%d%d",&a,&b)!=-1){
        bool flag;
        int mul=__gcd(a,b)*b,x,y;
        if(a*a-4*mul<0){
            flag=0;
        }else{
            int cha=sqrt(a*a-4*mul);
            x=(cha+a)/2;
            y=(a-cha)/2;
            if(x+y==a&&x/__gcd(x,y)*y==b){
                flag=1;
            }else{
                flag=0;
            }
        }
        if(!flag){
            printf("No Solution\n");
        }else{
            if(x>y) swap(x,y);
            printf("%d %d\n",x,y);
        }
    }

    return 0;
}