hdu 4627 The Unsolvable Problem【hdu2013多校3签到】

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本文介绍了一道关于寻找两个正整数的最大最小公倍数(LCM)的问题，给定一个整数N，需要找到两个正整数a和b使得a+b=N，并且[a,b]尽可能大，[a,b]表示a和b的最小公倍数。文章提供了详细的解题思路及C++实现代码。

链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4627

The Unsolvable Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 243 Accepted Submission(s): 143

Problem Description

There are many unsolvable problem in the world.It could be about one or about zero.But this time it is about bigger number.
Given an integer n(2 <= n <= 10 ⁹).We should find a pair of positive integer a, b so that a + b = n and [a, b] is as large as possible. [a, b] denote the least common multiplier of a, b.

Input

The first line contains integer T(1<= T<= 10000),denote the number of the test cases.
For each test cases,the first line contains an integer n.

Output

For each test cases, print the maximum [a,b] in a line.

Sample Input

Sample Output

Source

2013 Multi-University Training Contest 3

题意：

找最大的最小公倍数

给你一个数 N ， a+b = N ,找最大的 lcm(a,b）

思路：

奇数的随便就看出来了，取一半就好了

偶数的，写个暴力程序，打下表也可以随便看出规律，取一半了

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;

__int64 gcd(__int64 a, __int64 b)
{
    return b == 0 ? a : gcd(b,a%b);
}

__int64 lcm(__int64 a, __int64 b){
    return a/gcd(a,b)*b;
}

int main()
{
    int T;
    __int64 n;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%I64d", &n);
        __int64 a,b;
        __int64 ans = 0,tmp1,tmp2;
        if(n&1) ans = lcm(n/2,n/2+1);
        else  
        {
            if(n == 2) ans = 1;
            else
            {
                __int64 c = n/2-1;
                tmp1 = lcm(c,n-c);
                tmp2 = lcm(c-1,(n-c+1));
                ans = max(tmp1,tmp2);
            }
        }
        printf("%I64d\n", ans);
    }
    return 0;
}