2016 ICPC大连赛区 hdu5974 A Simple Math Problem

A Simple Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 456    Accepted Submission(s): 179

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
  

 

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:   5981  5980  5979  5978  5977 

题意：意思很简单，给你a和b，问能否找出两个数x，y，符合x+y=a，lcm（x，y）=b。

思路：其实就是一道公式题。。设k=gcd（x，y），a=k（x+y），b=kxy，很明显gcd（a，b）=k（下面会给出证明），所以x+y=a/k，xy=b/k，然后就是解二元一次方程，直接套公式就可以得到。然后这道题没说清楚，输出的时候x要比y小，不然是过不了的。下面给代码：

关于gcd(x+y,lcm(x,y))==gcd(x,y)==g的证明

令

x=g*k1;

y=g*k2

易知gcd(k1,k2)==1

所以x+y==g*(k1+k2)    lcm(x,y)==g*k1*k2

那么gcd(x+y,lcm(x,y))==gcd(g*(k1+k2),g*k1*k2)

要证明gcd(g*(k1+k2),g*k1*k2)==g 等价于证明gcd(k1+k2,k1*k2)==1

我们知道 a,b互质的时候 gcd(a+b,a*b)肯定是互质的,因为a+b的质因数肯定不包含a和b。

a*b的质因数肯定只包含a和b。

那么gcd(a+b,a*b)==1

因为k1,k2是互质的，所以gcd(k1+k2,k1*k2)==1

所以gcd(g*(k1+k2),g*k1*k2)==g

又因为gcd(x,y)==gcd(g*k1,g*k2)==g

所以gcd(x,y)==gcd(g*(k1+k2),g*k1*k2)==gcd(x+y,lcm(x,y))==g

得证

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
typedef long long LL;
using namespace std;
#define maxn 1005
#define ll l,mid,now<<1
#define rr mid+1,r,now<<1|1
#define lson l1,mid,l2,r2,now<<1
#define rson mid+1,r1,l2,r2,now<<1|1
LL gcd(LL a, LL b){
	return !b ? a : gcd(b, a%b);
}
int main(){
	LL a, b;
	while (~scanf("%lld%lld", &a, &b)){
		LL k = gcd(a, b);
		LL num = a*a - 4 * b*k;
		LL sqnum = sqrt(num);
		//cout << num << endl << sqnum << endl;
		LL x1 = a - sqnum;
		LL x2 = a + sqnum;
		if (sqnum*sqnum != num || x1 % (k << 1) || x2 % (k << 1)){
			printf("No Solution\n");
		}
		else{
			printf("%lld %lld\n", k*x1 / (k << 1), k*x2 / (k << 1));
		}
	}
}