https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5748

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 ∗ 104 ), b (1 ≤ b ≤ 109 ), and their meanings are shown in the description.

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

題目大意：

給定a,b，找到一組x,y滿足

x + y = a；

lcm（x，y）=b；

化簡式子

y = x - a;

x *  y = b * gcd(x,y);

設k=gcd(x,y) 可得 x * (x-a) = k * b;

那麼，我們可以通過列舉k得到x，並驗證k是否等於gcd(x,y)。

列舉k時嚶判斷求得x是否為整數。

#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

long long gcd(long long x,long long y)
{
     return y?gcd(y,x%y):x;
}

int main()
{
     long long a,b;
     long long x,y;
     bool found;

     while(~scanf("%lld%lld",&a,&b)){
          found=0;
          long long sz=min((a*a)/4/b+1,a);
          for(long long k=1;k<sz;++k){
               double c=sqrt(a*a-4*k*b);
               if(!(c-(int(c)))&&!((a+(int)c)&1)){
                    x=(a-c)/2;y=(a+c)/2;
                    if(gcd(x,y)==k){
                         found=1;
                         break;
                    }
               }
          }
          if(found)
               printf("%lld %lld\n",x,y);
          else
               printf("No Solution\n");
     }

     return 0;
}