HDU 5974 A Simple math problem (math problem)

Problem Description

Given positive integers A and b,find suitable X and Y to meet the conditions:
                                                        X+y=a
                                              Least Common Multiple (X, Y) =b

Inputinput includes multiple sets of test data. Each test data occupies one line,including, positive integers a (1≤a≤2*10^4), B (1≤b≤10^9), and their meanings is shown in The description. Contains most of the 12W test cases.

Outputfor each set of input data,output a line of the integers,representing x Y.if you cannot find such x and Y,output on E Line of "No solution" (without quotation) .  test instructions: give you two number a, if you can find two numbers x, y to make x+y=a and X, y least common multiple to B.   Because the topic gives the number of data groups, more than 12w groups so the traversal will time out, so try to find a way to evaluate directly. According to test instructions, two formulas can be obtained: (1) x+y=a; (2) X*y=b*k; (k is x, y greatest common divisor) obviously there is a k here is not good for the answer, so try to get rid of K. Will (1) all except K to get (3) x/k+y/k=a/k, and then (2) all except K to get (4) (x/k) * (y/k) =b/k; because k is x, y greatest common divisor, x/k, and y/k coprime. So a/k, and b/k is also coprime. So the problem is simplified to I+J=A/GCD (A, B), I*J=B/GCD (A, B), and the solutions of I and J.  

 #include <iostream> #include <cmath>using namespace Std;int X, y;int gcd (int a, int b) {return b > 0 gcd (b, a% B): A;}    int cal (int a, int b, int c) {if (A * a-4 * b < 0) return 0;    int GG = A * a-4 * b;    int FFF = sqrt (GG);        if (GG! = FFF * FFF) {Y =-1, X =-1;    return 0;    } X = (A + FFF);    Y = (A-FFF);        if (x% 2 = = 0 && Y% 2 = = 0) {X/= 2;        Y/= 2;    return 1;        } else {X =-1;        Y =-1;    return 0;    }}int Main () {int A, B;        while (Cin >> a >> b) {int c = gcd (A, b);        X =-1, Y =-1; if (Cal (A/C, b/c, c) && X! =-1 && Y! =-1) {cout << min (X, Y) * C << ' &L        t;< Max (X, Y) * C << Endl;        } else {cout << "No solution" << Endl; }} return 0;} 

Hdu 5974 A Simple math problem (math problem)