hdu 5974 A Simple Math Problem

来源：互联网发布：酒店门锁软件编辑：程序博客网时间：2024/05/16 06:14

A Simple Math Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 70 Accepted Submission(s): 39

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）

这是大连区域赛上的D题，当时是我队友做的，现在重现赛上写一下。
话说最近一直看别人的代码，好久没写代码了，T T。

思路：
由题意可知：
             lcm(x,y)=b;
因为         x*y =gcd(x,y)*lcm(x,y);
所以           x*y =gcd(x,y)*b;
因为             gcd(a,b)=gcd(x,y);
所以             x*y=b*gcd(a,b);
联立           x+y=a;

解个方程就行；

代码：

#include <cstdio>#include <cstring>#include <cmath>int gcd (int m,int n){int flag;while(n>0){flag=m%n;m=n;n=flag;}return m;}int main (){int a,b;while(scanf("%d%d",&a,&b)!=EOF){int i,j;int c=gcd(a,b);int x,y;int delta= a*a-4*b*c;if(delta<0){printf("No Solution\n");continue;}int flag= sqrt(delta);if(flag*flag!=delta||(a-flag)%2!=0){printf("No Solution\n");continue;}x=(1*a-sqrt(delta))/2;y=a-x;printf("%d %d\n",x<y?x:y,x<y?y:x);}return 0;}

0 0