uva 993 - Product of digits-优快云博客

本文链接：https://blog.youkuaiyun.com/yanshiqin4/article/details/20068779

本文介绍了一种算法，用于寻找给定非负整数N时，使得所有数字Q的乘积等于N的最小自然数Q。若不存在这样的Q，则返回-1。通过分解质因数并重新组合的方式找到解决方案。

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Product of digits

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 .

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

方法 ：

如果 n < 10,直接返回n

否则 从9到2计算n中的最大次幂数，n = 9^x8^y...2^z

然后计算n按因子升序将每个因子幂的个数展开的值，即为所求。

代码：



#include 
int min_num(int n)
{
    int i, t, digit[10];
    for(i = 0; i < 10; i++)
        digit[i] = 0;
    if(n < 10)
       return n;
    for(i = 9; i > 1; i--){
        while(n > 1 && (n % i) == 0){
            digit[i]++;
            n /= i;
        }
    }
    if(n > 9)
        return -1;
    t = 0;
    for(i = 2; i < 10; i++){
        while(digit[i]){
            t = t * 10 + i;
            digit[i]--;
        }
    }
    return t;
}
int main()
{
    int t, n;
    scanf("%d", &t);
    while(t-- > 0){
        scanf("%d", &n);
        printf("%d\n", min_num(n));
    }
    return 0;
}