HDU

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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): 1567 Accepted Submission(s): 434

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 8798 10780

Sample Output

No Solution308 490

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）

设gcd（x，y）= k

x=i*k，y=j*k，且i，j互质

则a=（i+j）*k

b=x*y/gcd（x，y）= i*j*k

i，j互质，i+j和i*j也互质（因为i*j的因子只有i，j和1，而i+j的因子是绝对不会有i和j的，可以用反证法证明）

因此gcd（a，b）=k==gcd（x，y）

这样基本就可解了

x+y=a；

x*y=b*k；

#include<bits/stdc++.h>using namespace std;long long gcd(long long a,long long b){return b?gcd(b,a%b):a;}int main(){long long a,b;while(~scanf("%lld %lld",&a,&b)){long long l=gcd(a,b);long long xx=a*a-4*b*l;if(xx<0)printf("No Solution\n");else{long long ans1=(long long)((a+sqrt(xx))/2);long long ans2=a-ans1;if(ans1*ans2/gcd(ans1,ans2)==b)printf("%lld %lld\n",ans2,ans1);elseprintf("No Solution\n");}}return 0;}

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