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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): 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 }

HDU-5974

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原文地址：http://www.cnblogs.com/Kiven5197/p/7875273.html