hdu 5974 A Simple Math Problem

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
Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）

题意：给定a，b,找到一组数据x,y使得 x+y=a,lcm(x,y)=b；

思路：学弟猜测：gcd(x,y)=gcd(a,b);

严格证明（转自:点我啊）:

已知条件：

①x+y=a,

②lcm(x,y)=b.

令gcd(x,y)=g；

则 x/g=k1 ,y/g=k2 且 k1与k2互质 (由于gcd为最大公约数)；

==>  g*k1+g*k2=a;

         g*k1*k2=b;

==>  k1+k2=a/g;

         k1*k2=b/g;

因k1与k2互质，所以k1+k2与k1*k2互质；

所以 a/g 与 b/g 也互质！

那么g就是gcd(a,b)；

所以gcd(x,y)=gcd(a,b)；

代码:

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int main(){    int a,b,c;    while(scanf("%d%d",&a,&b)!=EOF)    {        c=__gcd(a,b);        int x,y;        if((a*a-4*b*c)<0)        {            printf("No Solution\n");            continue;        }        x=(int)((a+sqrt(a*a-4*b*c))/2);        y=a-x;        if(x*y/__gcd(x,y)!=b)        {            printf("No Solution\n");            continue;        }        if(x>y)            swap(x,y);        printf("%d %d\n",x,y);    }    return 0;}