2016 ACM/ICPC大连区域赛 D—A Simple Math Problem【LCM+数学方程】

http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1004&cid=736

A Simple Math Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 729    Accepted Submission(s): 177
 

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

题意：给你两个数a,b，让你拆成x+y=a且lcm(x,y)=b。（x,y都是整数）

求x,y

分析：

令x=ck,y=dk，则b=cdk(k=gcd(x,y))

即可得

ck+dk=a

cdk=b

解方程。

隐含条件：输出的时候小的在前，大的在后。

代码：

#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f3f3f3f3fLL
using namespace std;
const int maxn=100010;
const ll mo=1e9+7;
ll a,b;
ll gcd(ll x,ll y){return y==0?x:gcd(y,x%y);}
int main()
{
    while(scanf("%lld%lld",&a,&b)!=EOF)
    {
        ll k=gcd(a,b);
        ll dt=a*a-4*k*b;
        ll c=sqrt(dt);
        if(dt>=0&&c*c==dt)
        {
            ll tmp=(a+c)/(2*k);
            if((a+c)%(2*k)==0&&b%(tmp*k)==0)
            {
                ll cnt=b/(tmp*k);
                printf("%lld %lld\n",cnt*k,tmp*k);
            }
            else {
            tmp=(a-c)/(2*k);
            if((a-c)%(2*k)==0&&b%(tmp*k)==0)
            {
                ll cnt=b/(tmp*k);
                printf("%lld %lld\n",cnt*k,tmp*k);
            }
            else puts("No Solution");
            }
        }
        else puts("No Solution");
    }
    return 0;
}