UVa 993 Product of digits(简单数论)_993分解因数-优快云博客

本文介绍UVA 993题目，该题要求找到最小的自然数Q，使得Q所有位上的数字之积等于给定的非负整数N。通过素因数分解的方法，实现O(logN)的复杂度解决此问题。

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993 - Product of digits

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=934

For a given non-negative integer number N , find the minimal natural Q such that the product of all digits of Q is equal N .

Input

The first line of input contains one positive integer number, which is the number of data sets. Each subsequent line contains one data set which consists of one non-negative integer number N (0 $\le$ N $\le$ 10⁹) .

Output

For each data set, write one line containing the corresponding natural number Q or `-1' if Q does not exist.

Sample Input

Sample Output

1 
25 
-1

题意：

给一个数N，是否存在这样的数Q，Q的所有位上的数之积等于N？

思路：

将N素因数分解，再合并整理即可。

复杂度：O(log N)

完整代码：

/*0.012s*/

#include <cstdio>
#include <cstring>

int main()
{
	int t, index[8];
	scanf("%d", &t);
flag:
	while (t--)
	{
		memset(index, 0, sizeof(index));
		int n;
		scanf("%d", &n);
		if (n <= 1)
		{
			printf("%d\n", n);
			continue;
		}
		//分解N
		while (n != 1)
		{
			int count = 0;
			if (!(n & 1)) index[0]++, n >>= 1;
			else count++;
			if (!(n % 3)) index[1]++, n /= 3;
			else count++;
			if (!(n % 5)) index[3]++, n /= 5;
			else count++;
			if (!(n % 7)) index[5]++, n /= 7;
			else count++;
			if (count == 4)
			{
				puts("-1");
				goto flag;
			}
		}
		//2^3=8
		index[6] += index[0] / 3, index[0] %= 3;
		//2^2=4
		if (index[0] == 2) index[2] = 1, index[0] = 0;
		//3^2=9
		index[7] += index[1] >> 1, index[1] &= 1;
		//2*3=6
		if (index[0] & index[1]) index[4] = 1, index[1] = index[0] = 0;
		for (int i = 0; i < 8; ++i)
			while (index[i]--) putchar(i + 50);
		putchar(10);
	}
	return 0;
}