hdu5974

Posted 2020-09-18 不积跬步无以至千里，不积小流无以成江海

A Simple Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1795    Accepted Submission(s): 524

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

 1 #include <iostream>
 2 #include <cmath>
 3 #include <algorithm>
 4 #include <string>
 5 #include <cstring>
 6 #include <cstdio>
 7 using namespace std;
 8 long long f(long long a,long long b)
 9 {
10     return (b==0)?a:f(b,a%b);
11 }
12 int main()
13 {
14     long long a,b,l,ans,t;
15     while(~scanf("%I64d%I64d",&a,&b))
16     {
17         l=f(a,b);
18         if(a*a-4*b*l<0)
19             cout<<"No Solution"<<endl;
20         else
21         {
22             ans=(a-sqrt(a*a-4*b*l))/2;
23             t=a-ans;
24             if(ans*t/f(t,ans)==b)
25                 cout<<ans<<‘ ‘<<t<<endl;
26             else
27               cout<<"No Solution"<<endl;
28         }
29     }
30     return 0;
31 
32 }