CodeForces 1285 C Fadi and LCM （GCD）

Description：

Today, Osama gave Fadi an integer $X$ , and Fadi was wondering about the minimum possible value of $max(a,b)$ such that $LCM(a,b)$ equals $X$ . Both $a$ and $b$ should be positive integers.

$LCM(a,b)$ is the smallest positive integer that is divisible by both $a$ and $b$ . For example, $LCM(6,8)=24, LCM(4,12)=12, LCM(2,3)=6.$

Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair?

Input

The first and only line contains an integer $X (1≤X≤10^12).$

Output

Print two positive integers, $a$ and $b$ , such that the value of $max(a,b)$ is minimum possible and $LCM(a,b)$ equals $X$ . If there are several possible such pairs, you can print any.

Examples

input

output

1 2

input

output

2 3

input

output

1 4

input

output

1 1

题意：

$\quad$ 给定一个数 $x$ ，要找出两个数 $a，b$ ，使得 $a，b$ 的最小公倍数等于 $x$ 。且 $a，b$ 中最大的值尽可能的小。即 $max(a,b)$ 尽可能小。
$\quad$ 易知当 $a*b=X$ 时 $max(a,b)$ 最小。且 $\frac{a * b} {gcd(a,b)}=lcm(a,b)$ 即 $a*b=lcm(a,b$ $gcd(a,b)=1$ ,所以直接枚举答案判断。

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

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

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
    if (a >= mod)
        a = a % mod + mod;
    ll ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = ans * a;
            if (ans >= mod)
                ans = ans % mod + mod;
        }
        a *= a;
        if (a >= mod)
            a = a % mod + mod;
        b >>= 1;
    }
    return ans;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}

///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //算出它们累乘的结果
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //计算逆元
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}

int t;
int n, m;
int ans, res, temp;
ll x;
ll ans1, ans2;

bool judge(ll num)
{
    if (num <= 3)
    {
        return num > 1;
    }
    if (num % 6 != 1 && num % 6 != 5)
    {
        return false;
    }
    int q = sqrt(num);
    for (ll i = 5; i <= q; i += 6)
    {
        if (num % i == 0 || num % (i + 2) == 0)
        {
            return false;
        }
    }
    return true;
}

int main()
{
    while (~sld(x))
    {
        if (judge(x))
        {
            cout << "1"
                 << " " << x << endl;
            continue;
        }
        ll q = sqrt(x);
        for (q; q > 0; q--)
        {
            if (x % q == 0)
            {
                if (gcd(q, x / q) == 1)
                {
                    ans1 = q;
                    ans2 = x / q;
                    break;
                }
            }
        }
        pldd(ans1, ans2);
    }
    return 0;
}