HDU-5974-A Simple Math Problem

Problem 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

题意：
已知 a,b，找出一个x 和 y 满足 x+y=a，Lcm（x,y）=b;

思路(上网查找)：
看数据范围肯定不能进行暴力枚举了！
令gcd(x,y) = g;
那么
g * k1 = x;
g * k2 = y;
因为g 是最大公约数，那么k1与k2 必互质！
=> g*k1*k2 = b
=> g*k1 + g * k2 = a;
所以k1 * k2 = b / g;
k1 + k2 = a/g;
因为k1与k2 互质！
所以k1 * k2 和 k1 + k2 也一定互质（一个新学的知识点= = ）
所以a/g 与b/g也互质！
那么g 就是gcd(a,b);
所以我们得出一个结论： gcd(x,y) == gcd(a,b);;
所以x + y 与 x * y都是已知的了，解一元二次方程即可！

因为 gcd(a,b) * lcm(a,b) = a*b;
所以 lcm(a,b) = a/gcd(a,b)*b;
所以题目条件可化为一个一元二次方程：x*x-a*x+b*gcd(a,b)。

代码

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>using namespace std;int a,b;int gcd(int m,int n){    if(n == 0)        return m;    else        return gcd(n,m%n);}int main(){    int m,p,l;    int q;    int x1,x2;    int flag;    while(scanf("%d%d",&a,&b)!=EOF)    {        flag = 1;        m = gcd(a,b);        p = a*a-4*b*m;  //解一元二次方程        if(p < 0)  //无解        {            printf("No Solution\n");            continue;        }        q = (int)sqrt(p);        if(q*q != p)//是否为整数            flag = 0;        x1 = (a+q)/2;        x2 = (a-q)/2;        if(flag == 0)            printf("No Solution\n");        else        {            l = min(x1,x2);            printf("%d %d\n",l,a-l);        }    }    return 0;}