Codeforces 588B Duff in Love 【数学】

最新推荐文章于 2021-11-08 18:03:27 发布

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最新推荐文章于 2021-11-08 18:03:27 发布

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分类专栏： codeforces 数学

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本文介绍了一种算法，用于找出给定整数n的最大Lovely因子。Lovely因子定义为不含任何完全平方数因子的数。通过分解n的所有质因子并相乘得到最终结果。

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题目链接：Codeforces 588B Duff in Love
B. Duff in Love
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Duff is in love with lovely numbers! A positive integer x is called lovely if and only if there is no such positive integer a > 1 such that a2 is a divisor of x.

Malek has a number store! In his store, he has only divisors of positive integer n (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.

Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.

Input
The first and only line of input contains one integer, n (1 ≤ n ≤ 1012).

Output
Print the answer in one line.

Examples
input
10
output
10
input
12
output
6
Note
In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn’t divisible by any perfect square, so 10 is lovely.

In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 22, so 12 is not lovely, while 6 is indeed lovely.

定义lovely数：不存在大于1的完全平方数是它的因子。
题意：给定一个n，问你n的因子里面最大的lovely 数。

思路：求出所有质因子，全部乘起来就是结果。
AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MOD = 1e4 + 7;
const int MAXN = 1e5 + 10;
void add(LL &x, LL y) { x += y; x %= MOD; }
LL p[100]; int k;
void getp(LL n) {
    k = 0;
    for(LL i = 2; i * i <= n; i++) {
        if(n % i == 0) {
            p[k++] = i;
            while(n % i == 0) {
                n /= i;
            }
        }
    }
    if(n > 1) p[k++] = n;
}
int main() {
    LL n;
    while(scanf("%lld", &n) != EOF) {
        getp(n); LL ans = 1LL;
        for(int i = 0; i < k; i++) {
            ans = ans * p[i];
        }
        printf("%lld\n", ans);
    }
    return 0;
}