HDU5974 A Simple Math Problem(求根公式)

最新推荐文章于 2019-05-14 15:56:22 发布

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本篇介绍了一道编程解决的数学题目，给定两个正整数a和b，寻找合适的X和Y，使得X+Y=a且X和Y的最小公倍数为b。文章提供了一个C++实现的解决方案，利用最大公约数和最小公倍数之间的关系，通过求根公式找到可能的X和Y值。

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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): 3691 Accepted Submission(s): 1156

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a,b,求满足X+Y=a,lcm(X,Y)=b,的X，Y

思路

设X，Y的最大公约数为t，即 $g c d (X, Y) = t$ ，那么有 $X=k_{1}t$ , $Y=k_{2}t$ ,其中 $k_{1},k_{2}$ 互质， $X+Y=(k_{1}+k_{2})t$ , $lcm(X,Y)=XYt=k1k2tlcm(X,Y)=\frac{XY}{t}=k_{1}k_{2}t$ ,又因为 $k_{1}+k_{2}$ 与 $k_{1}k_{2}$ 互质，所以 $g c d (X + Y, l c m (X, Y)) = t$ ,又有 $g c d (X, Y) = t$ ,所以 $g c d (X, Y) = g c d (X + Y, l c m (X, Y)) = g c d (a, b) = t$
然后又 $lcm(X,Y)=XYtlcm(X,Y)=\frac{XY}{t}$ , $Y = a - X$
得 $X^{2}-aX+tb=0$ ,其中 $t = g c d (a, b)$ ,用求根公式求解即可，X为较小一项，求完后再检验一下

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <math.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)!=EOF)
    {
        int t=gcd(a,b);
        if(a*a-4*t*b<0)
            printf("No Solution\n");
        else
        {
            long long x=min((a+sqrt(a*a-4*t*b))/2,(a-sqrt(a*a-4*t*b))/2);
            long long y=a-x;
            if(x*y/t==b)
            printf("%lld %lld\n",x,y);
            else
                printf("No Solution\n");
        }
    }
    return 0;
}