HDU 5974 A Simple Math Problem(简单数学)

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

题目大意

给定两个值 a,b，判断是否存在两个数 x,y 满足 x+y=a，lcm(x,y)=b，
若存在，输出x,y，否则输出“No Solution”。
注意：输出有顺序要求，必须是小值在前。

解题思路

某童鞋厉害得不行的直接看出了规律还是定理？
如果这两个值存在的话，就满足 gcd(x,y)=gcd(a,b)，然后可得 x*y = lcm(x,y)*gcd(x,y)=gcd(x,y)*b=gcd(a,b)*b，再由完全平方公式可以得到 x 与 y 的差值，再与 x+y 相加减就可以得到 x,y，最后记得验证求得的值x+y=a，lcm(x,y)=b是否成立。
最后推一下这个规律还是定理？
设x=m*gcd(x,y),y=n*gcd(x,y)；
gcd(a,b)=gcd(x+y,lcm(x,y))=gcd(m*gcd(x,y)+n*gcd(x,y),m*n*gcd(x,y))
             =gcd(x,y)*gcd(m+n,m*n)；
因为 m,n 是互质的，所以 m+n 与 m*n 也是互质的，即可得gcd(x,y)=gcd(a,b)。

代码实现

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;#define ll long longll a,b;ll gcd(ll a,ll b){    if(b==0) return a;    return gcd(b,a%b);}int main(){    while(~scanf("%lld %lld",&a,&b))    {        ll g=gcd(a,b);        ll xy=b*g;        ll z=a*a-4*xy;        if(z<0)        {            printf("No Solution\n");            continue;        }        else z=sqrt(z);        ll x=(z+a)/2;        ll y=a-x;        if(x*y/gcd(x,y)!=b||x+y!=a)            printf("No Solution\n");        else            printf("%lld %lld\n",y,x);    }    return 0;}