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：给定a，b，满足x+y=a、lcm（最小公倍数）（x，y）=b，求这样的x，y

思路：12w的测试数据，当然不能暴力。

首先肯定能想到x*y/gcd(x,y)==b;   {gcd(a,b)代表a，b的最大公约数}

设gcd（x，y）=t；

i*t = x；

j*t = y；

则i和j肯定互质；

所以i+j和i*j也肯定互质（下面会用到此规则，不明白为什么的，可记住该结论）

i*t + j*t = a;

i*j*t = b;

由i+j和i*j互质，所以gcd（a，b） = t；（解决该题的关键一波转换）

故i+j = a/t；

    i*j = b/t；

这马上联想到高中学的韦达定理；

构造二次函数t*x^2-a*t+b = 0;

用求根公式求出来即可得i ， j； （取正根）

当然中间还有一些判别式的判定。

代码如下：

#include<stdio.h>#include<string.h>#include<iostream>#include<math.h>using namespace std;int gcd(int a,int b){    if(b==0)        return a;    return gcd(b,a%b);}int main(){    int a,b,t,x,y;    while(~scanf("%d%d",&a,&b))    {        t = gcd(a,b);       //最大公约数        if(a*a-4*t*b<0)  //判别式判断是否有根            printf("No Solution\n");        else        {            int i = ((a+sqrt(a*a-4*t*b))/(2*t));            x = i*t;            y = a-x;            if(x*y/gcd(x,y)==b)  //判定解得的根是否正确            {                if(x<=y)                    printf("%d %d\n",x,y);                else                    printf("%d %d\n",y,x);            }            else                   printf("No Solution\n");        }    }    return 0;}