HDU-5974

A Simple Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3646    Accepted Submission(s): 1140


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

AC代码:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int MAXX=10010;
 5 
 6 int gcd(int a,int b){
 7     return (b>0)?gcd(b,a%b):a;
 8 }
 9 
10 int main(){
11     ios::sync_with_stdio(false);
12     long long a,b,x,y;
13     while(cin>>a>>b&&a&&b){
14         long long temp=a;
15         long long l=gcd(a,b);
16         a/=l;
17         b/=l;
18         if(a*a-4*b<0){
19             cout<<"No Solution"<<endl;
20         }
21         else{
22             y=(a+sqrt(a*a-4*b))/2;
23             y*=l;
24             x=temp-y;
25             if(x*y/gcd(x,y)==b*l){
26                 cout<<x<<" "<<y<<endl;
27             }
28             else{
29                 cout<<"No Solution"<<endl;
30             } 
31         }
32     }
33     return 0;
34 }
原文地址:https://www.cnblogs.com/Kiven5197/p/7875273.html