HDU 5974 A Simple Math Problem 数学

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A Simple Math Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2505 Accepted Submission(s): 777

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

Sample Output

  
  
   
   No Solution
308 490

一道纯的数学题。

题意：给出两个数 a,b。a是x和y的和，b是x和y的最小公倍数；

在这里根据两个数互质，则他们的和与他们的积也互质可以推导出a和b的最大公因数就是x和y的最大公因数。

然后解方程，在这里注意一下，求得的解必须是整数。

#include<stdio.h>
#include<math.h>
#define ll long long
using namespace std;
ll gcd(ll a,ll b)
{   while(b!=0)
    {ll c=a%b;
         a=b;
         b=c;

    }
    return a;
}
ll lcm(ll a,ll b)
{
    return a/gcd(a,b)*b;
}
int main()
{
    ll a,b;
    while(scanf("%lld%lld",&a,&b)!=EOF)
    {
        ll c=gcd(a,b);
        c=b*c;
        ll d=a-sqrt(a*a-4*c);
        ll p=d/2;
        if(a*a-4*c<0)
            printf("No Solution\n");
        else
        {
            ll z=a*a-4*c;
            ll s=sqrt(z);
            if(s*s==z&&p*2==d)
                printf("%lld %lld\n",p,a-p);
            else
                printf("No Solution\n");
        }

    }
}