Codeforces ~ 1062B ~ Math(思维,因数分解)

本文介绍了一种算法,用于计算将一个给定数字通过乘法和开方操作最小化的最少步骤。通过分解质因数并分析素因子的幂,算法确定了达到最小数字所需的最优操作次数。

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在这里插入图片描述

题意

给你一个数,现在有两种操作:
①将这个数字乘以x
②将x开根号
问这个数字最小能变成的数字是多少,最少需要多少次操作

思路

把n分解质因数以后。开方不能减少素因子的种类,但是能够减少每个素因子的幂,每次开方相当于幂/2。
所以最小数一定是每种素因子相乘,最小操作次数就是开方次数+乘法次数。
如果所有幂值都是222的同一个次方,接下来就是一直幂/2了,否则我们就要先通过一次乘法将所有幂值变为2的同一个次方。

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5+5;
int n;
vector<pair<int, int> > v;
void get_f(int x)
{
    for (int i = 2; i*i <= x; i++)
    {
        int cnt = 0;
        while (x%i == 0) x /= i, cnt++;
        if (cnt) v.emplace_back(i, cnt);
    }
    if (x > 1) v.emplace_back(x, 1);
}
int main()
{
    scanf("%d", &n);
    get_f(n);
    int ans = 1, MAX = 0, flag = false;
    for (int i = 0; i < v.size(); i++)
    {
        int a = v[i].first, b = v[i].second;
        //cout << a << "," << b << endl;
        if ((i > 0 && v[i-1].second != b) || (b&(b-1)) != 0 || b == 1) flag = true;
        ans *= a;
        MAX = max(MAX, b);
    }
    int cnt = 0;
    if (MAX > 1)
    {
        cnt += (int)ceil(log2(MAX));
        cnt += flag;
    }
    printf("%d %d\n", ans, cnt);
    return 0;
}
/*
5184
*/

翻译:# CF1444A Division ## 题目描述 Oleg's favorite subjects are History and Math, and his favorite branch of mathematics is division. To improve his division skills, Oleg came up with $ t $ pairs of integers $ p_i $ and $ q_i $ and for each pair decided to find the greatest integer $ x_i $ , such that: - $ p_i $ is divisible by $ x_i $ ; - $ x_i $ is not divisible by $ q_i $ . Oleg is really good at division and managed to find all the answers quickly, how about you? ## 输入格式 The first line contains an integer $ t $ ( $ 1 \le t \le 50 $ ) — the number of pairs. Each of the following $ t $ lines contains two integers $ p_i $ and $ q_i $ ( $ 1 \le p_i \le 10^{18} $ ; $ 2 \le q_i \le 10^{9} $ ) — the $ i $ -th pair of integers. ## 输出格式 Print $ t $ integers: the $ i $ -th integer is the largest $ x_i $ such that $ p_i $ is divisible by $ x_i $ , but $ x_i $ is not divisible by $ q_i $ . One can show that there is always at least one value of $ x_i $ satisfying the divisibility conditions for the given constraints. ## 输入输出样例 #1 ### 输入 #1 ``` 3 10 4 12 6 179 822 ``` ### 输出 #1 ``` 10 4 179 ``` ## 说明/提示 For the first pair, where $ p_1 = 10 $ and $ q_1 = 4 $ , the answer is $ x_1 = 10 $ , since it is the greatest divisor of $ 10 $ and $ 10 $ is not divisible by $ 4 $ . For the second pair, where $ p_2 = 12 $ and $ q_2 = 6 $ , note that - $ 12 $ is not a valid $ x_2 $ , since $ 12 $ is divisible by $ q_2 = 6 $ ; - $ 6 $ is not valid $ x_2 $ as well: $ 6 $ is also divisible by $ q_2 = 6 $ . The next available divisor of $ p_2 = 12 $ is $ 4 $ , which is the answer, since $ 4 $ is not divisible by $ 6 $ .
最新发布
07-11
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