Codeforces Round #565 (Div. 3)A. Divide it!

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最新推荐文章于 2024-07-25 20:12:28 发布

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分类专栏： codeforces 文章标签： C/C++

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本文探讨了一个算法问题，目标是最小化操作次数将任意正整数n转换为1。允许的操作包括：若n能被2整除，则用n/2替换n；若n能被3整除，则用2*(n/3)替换n；若n能被5整除，则用4*(n/5)替换n。文章详细解释了如何通过这些操作达到目标，并提供了一个实例演示。

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You are given an integer n.

You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:

Replace n with n2 if n is divisible by 2;
Replace n with 2n3 if n is divisible by 3;
Replace n with 4n5 if n is divisible by 5.
For example, you can replace 30 with 15 using the first operation, with 20 using the second operation or with 24 using the third operation.

Your task is to find the minimum number of moves required to obtain 1 from n or say that it is impossible to do it.

You have to answer q independent queries.

Input
The first line of the input contains one integer q (1≤q≤1000) — the number of queries.

The next q lines contain the queries. For each query you are given the integer number n (1≤n≤1018).

Output
Print the answer for each query on a new line. If it is impossible to obtain 1 from n, print -1. Otherwise, print the minimum number of moves required to do it.

Example
inputCopy
7
1
10
25
30
14
27
1000000000000000000
outputCopy
0
4
6
6
-1
6
72

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<string>
#define ll long long
#define dd double
using namespace std;

int main() {
	ios::sync_with_stdio(0);
	ll t;
	cin >> t;
	while (t--) {
		ll n; cin >> n;
		ll sum = 0;
		ll flag = 0;
		while (n != 1) {
			if (n % 2 == 0) {
				n = n / 2;
				sum++;
			}
			else if(n % 3 == 0 && n % 5 == 0) {
				n = min(((n / 3) * 2),((n / 5) * 4));
				sum++;
			}
			else if (n % 3 == 0) {
				n = n / 3 * 2;
				sum++;
			}
			else if (n % 5 == 0) {
				n = n / 5 * 4;
				sum++;
			}
			else {
				flag = 1;
				cout << "-1" << endl;
				break;
			}
		}
		if (flag == 0) {
			cout << sum << endl;
		}
	}
}