Given two positive integers a and b,find suitable X and Y to meet the conditions:X+Y=aLeast 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

輸入兩個整數a,b,知否可以找到兩個整數x,y.使其滿足x+y=a,lcm(x,y)=b 如果找不到輸出No Solution 否則輸出x y(x<=y)

gcd(x,y)=gcd(a,b);

可以記住一個結論：a和b兩個數，可以分解成x,y使得x+y=a,x和y的最小公倍數是b,(lcm(x,y)=b, 那么有gcd(a,b)=gcd(x,y);

#include <iostream> #include <math.h> using namespace std; int gcd(int x,int y) {if(y==0)return x;elsereturn gcd(y,x%y); } int main() {int a,b,temp,k,d;while(cin>>a>>b){temp=gcd(a,b);b*=temp;k=a*a-4*b;d=sqrt(k);if(k<0||d*d!=k||(a+d)%2!=0){cout<<"No Solution"<<endl;continue;}d=(a+d)/2;k=a-d;if(d<=k)cout<<d<<' '<<k<<endl;elsecout<<k<<' '<<d<<endl;}return 0; }

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