HDU - 5974 A Simple Math Problem

A - A Simple Math Problem

HDU - 5974

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

InputInput 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.OutputFor 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

OJ-ID：
hdu-5974

author：
Caution_X

date of submission：
20191028

tags：
math

description modelling：
输入a,b，问是否存在x,y满足x+y=a && lcm(x,y)=b，若存在，输出一组(x,y)

major steps to solve it：
数论知识铺垫：
若i,j互质，则i+j和i*j互质
证明：
假设i,j互质但是i+j和i*j不互质，则(i+j)*k=i*j----①
两边同时%i得j*k%i=0,因为i，j互质，假如①能成立，你那么k%i一定为0
假设k%i=0，设k=m*i，k代入①得i*i*m+m*i*j=i*j，两边同时%j得到i*i*m%j=0
因为i，j互质，则m%j=0，设m=h*j (h>1)，则k=h*i*j
那么①：(i+j)*h*i*j=i*j   =>  (i+j)*h=1，显然该等式不成立，即k%i=0不成立
又因为k%i=0不成立，所以(i+j)*k=i*j不成立，所以(i+j)和i*j互质。
本题分析：
设g=gcd(x,y)，i=x/g,j=y/g，
那么i*j*g=lcm(x,y)=b, (i+j)*g=a
因为i,j互质，则i+j和i*j互质，gcd(a,b)=gcd((i+j)*g,i*j*g)=g=gcd(x,y)  => gcd(a,b)=gcd(x,y)---②
由x+y=a,lcm(x,y)=b,gcd(a,b)=gcd(x,y)三式联立求解一元二次方程即可

warnings：
输出的x,y满足x<y（题目没有明说）若输出结果为x,y，x,y符合条件但是x>y,会WA.

AC code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll gcd(ll x,ll y)
{
    if(y == 0)
        return x;
    else
        return gcd(y,x%y);
}
ll big(ll a,ll b,ll l)
{
    return a*b/l;
}
int main()
{
    ll a,b;
    while(~scanf("%lld %lld",&a,&b))
    {
        ll g = gcd(a,b);
        ll bb = a/g;
        bb = -bb;
        ll c = b/g;
        if(bb*bb>=4*c)
        {
            ll x = (ll)(-bb+sqrt(bb*bb-4*c))/2;
            ll y = -bb-x;
            ll xx = x*g;
            ll yy = y*g;
            if(xx>yy)
                swap(xx,yy);
            if(xx+yy == a && big(xx,yy,g) == b)
                printf("%lld %lld\n",xx,yy);
            else
                printf("No Solution\n");
        }
        else
            printf("No Solution\n");
    }
    return 0;
}