2019长安大学ACM校赛网络同步赛 M LCM （数论）

链接：https://ac.nowcoder.com/acm/contest/897/M
来源：牛客网

LCM

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述

    Silly Slp knows nothing about number theory. One day he feels puzzled with the following problem.
    Give two positive integers n and c. Find a pair of positive integer (a, b), which satisfies both of a and b are no more than n and the lowest common multiple of them is c. Moreover, maximize $\mathrm{a\times ba\times b,\; the\; product\; of\; a\; and\; b.\; Please\; tell\; Silly\; Slp\; the\; maximize\; value\; of\; a\times ba\times b.\; If\; a\; and\; b\; don\text{'}t\; exist,\; print\; “-1”(without\; quotes).}$

输入描述:

    The first line contains an integer number T, the number of test cases.
    ${\mathrm{ithith\; of\; each\; next\; T\; lines\; contains\; two\; numbers\; n\; and\; c(1\le n,c\le 1061\le n,c\le 106).}}^{}$

输出描述:

For each test case print a number, the maximize value.

示例1

输入

复制

2
10 12
5 36

输出

复制

24
-1


题意：
给你一个n和一个c，
让你找两个数a，b，满足 a<=n,b<=n, lcm(a,b)=c
思路：
c=lcm(a,b)=a*b/gcd(a,b);
so, gcd(a,b) 是c的一个因子，
那么我们sqrt(c)的时间复杂度来枚举 c的因子i，我们假定i就是 gcd ，然后y=c/i
y=a*b/gcd(a,b)/gcd(a,b)
我们再来枚举 y的因子，如果y有两个因子 p，q， 满足gcd(p,q)==1 并且 p*gcd<=n,q*gcd<=n 
那么 p，q就可以是 a/gcd(a,b), b/gcd(a,b)的一种可能，
p*gcd*q*gcd = a*b 
我们枚举的时候取最大值，就可以得到我们想要的答案。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d
",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/

int main()
{
    // freopen("D:\common_text\code_stream\in.txt","r",stdin);
    //freopen("D:\common_text\code_stream\out.txt","w",stdout);
    int t;
    // gbtb;
    // cin>>t;
    scanf("%d", &t);
    while (t--)
    {
        ll n, c;
        scanf("%lld %lld", &n, &c);
        // cin>>n>>c;
        ll res = -1;
        for (ll i = 1; i * i <= c; i++)
        {
            if (c % i == 0)
            {

                ll y = c / i; //枚举gcd(a,b)=i  a*b=c*i
                for (ll j = 1; j * j <= y; j++) {
                    if (y % j == 0 && gcd(j, y / j) == 1 && j * i <= n && y / j * i <= n) {
                        res = max(res, c * i);
                    }
                }
                y = i; //枚举gcd(a,b)=c/i  a*b=c*(c/i)
                for (ll j = 1; j * j <= y; j++) {
                    if (y % j == 0 && gcd(j, y / j) == 1 && j * (c / i) <= n && y / j * (c / i) <= n) {
                        res = max(res, c * (c / i));
                    }
                }
            }
        }
        printf("%lld
", res);
        // cout<<ans<<endl;
    }


    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '
');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}