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 a×ba×b, the product of a and b.
    Please tell Silly Slp the maximize value of a×ba×b. If a and b don't exist, print “-1”(without quotes).

输入描述:

    The first line contains an integer number T, the number of test cases.
    ithith of each next T lines contains two numbers n and c(1≤n,c≤1061≤n,c≤106).

输出描述:

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

示例1

输入

复制

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\n",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), '\0', 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 = ; while (b) {if (b % )ans = ans * a % MOD; a = a * a % MOD; b /= ;} return ans;}

inline void getInt(int* p);

const int maxn = ;

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 = -;

        for (ll i = ; i * i <= c; i++)

        {

            if (c % i == )

            {

                ll y = c / i; //枚举gcd(a,b)=i  a*b=c*i

                for (ll j = ; j * j <= y; j++) {

                    if (y % j ==  && gcd(j, y / j) ==  && 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 = ; j * j <= y; j++) {

                    if (y % j ==  && gcd(j, y / j) ==  && j * (c / i) <= n && y / j * (c / i) <= n) {

                        res = max(res, c * (c / i));

                    }

                }

            }

        }

        printf("%lld\n", res);

        // cout<<ans<<endl;

    }

    return ;

}

inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '');

        while ((ch = getchar()) >= '' && ch <= '') {

            *p = *p *  - ch + '';

        }

    }

    else {

        *p = ch - '';

        while ((ch = getchar()) >= '' && ch <= '') {

            *p = *p *  + ch - '';

        }

    }

}