我在Glassdoor遇到了这个问题并试图实现它。问题如下 -
考虑一个数字 123，该数字与其数字的乘积 (123*1*2*3 = 738) 是 738。因此，123 是 738 的种子根。编写程序接受一个数字并找到所有可能的种子根. 例如，如果用户输入 4977，那么答案应该是 79 和 711。
我想到了一种方法：
找出 2 到 9 中除数的所有数字。
然后从最大的数字（在步骤 1 中找到的数字中）开始，找到组成数字的数字，然后打印这些数字的所有排列。
但是，这假设数字不重复，其次它不会像 4977 那样打印所有数字，它可以找到 79 但找不到 711。
有更好的方法吗？
Another solution is to check every divisor of n and see if this could be a seed. To check all the divisors, you just need to check to the square root of n. So this algorithm runs in O(sqrt(n)). Could you do it faster?
Here is a simple C++ programme demonstrating this idea.
#include<iostream>
using namespace std;
int prod(int n){
int res=n;
while(n!=0){
res*=n%10;
n/=10;
}
return res;
}
int main(){
int n;
cin >> n;
for(int i=1;i*i<=n;++i){
if(n%i==0){
if(prod(i)==n)
cout << i << endl;
if(n/i!=i && prod(n/i)==n)
cout << n/i << endl;
}
}
}
我的方法是这样的。这是一种递归算法，它使用集合 S，它是一个包含从 2 到 9 的数字的多重集，可能是多次。
try (N, S, digit) {
for d = digit, digit-1, ..., 2 {
if N is divisible by d then {
S' = S + {d};
if N/d is composed of all the digits in S', perhaps with
some 1's thrown in, then N/d is an answer;
try (N/d, S', d);
}
}
}
然后为原始号码
try (originalN, empty-set, 9);
also check originalN to see if it has only 1 digits (11, 11111, etc.); if so, then it's also an answer
我认为这会起作用，但我可能错过了一些边界情况。
对于 4977，try(4977, empty, 9)会发现 4977 可以被 9 整除，所以它调用try(553, {9}, 9). 内部try发现 553 可以被 7 整除，并且 553/7 = 79；此时 S' = {7, 9}，算法检查 79 是否由数字 {7, 9} 组成，结果成功。不过，该算法仍在继续。最终我们会回溯到外部try，它会在某个时候尝试d = 7，并且 4977/7 = 711，当我们进行检查时，S' = {7} 和 711 由 7 和一些 1 组成，所以这也是一个回答。
编辑：我已经包含了一个完整的 C++ 函数：
#include <iostream>
#include <vector>
using namespace std;
struct digitMultiset {
int counts[10]; // note: the [0] and [1] elements are not used
};
static const digitMultiset empty = { {0, 0, 0, 0, 0, 0, 0, 0, 0, 0} };
bool hasDigits (const int n, digitMultiset& s) {
digitMultiset s2 = empty;
int temp = n;
int digit;
while (temp > 0) {
digit = temp % 10;
if (digit == 0) return false;
if (digit != 1) s2.counts[digit]++;
temp /= 10;
}
for (int d = 2; d < 10; d++)
if (s2.counts[d] != s.counts[d]) return false;
return true;
}
void tryIt (const int currval, const digitMultiset& s,
const int digit, vector<int>& answers) {
digitMultiset newS;
for (int d = digit; d >= 2; d--) {
if (currval % d == 0) {
int quotient = currval / d;
newS = s;
newS.counts[d]++;
if (hasDigits (quotient, newS))
answers.push_back (quotient);
tryIt (quotient, newS, d, answers);
}
}
}
void seedProduct (const int val) {
vector<int> answers;
tryIt (val, empty, 9, answers);
int temp = val;
bool allOnes = true;
while (temp > 0) {
if (temp % 10 != 1) {
allOnes = false;
break;
}
temp /= 10;
}
if (allOnes)
answers.push_back(val);
int count = answers.size();
if (count > 0) {
if (count == 1)
cout << val << " has seed product " << answers[0] << endl;
else {
cout << val << " has " << count << " seed products: ";
for (int& ans : answers)
cout << ans << " ";
cout << endl;
}
}
}
176852740608 不到 1 秒。找到所有因素，然后检查特定因素是否是种子根。
import java.util.ArrayList;
public class Number {
long value;
public ArrayList<Long> factors= new ArrayList<Long>();
public Number(long a){
value = a;
for(long i=1;i<= Math.sqrt(a);i++){
if(a%i==0)
{
factors.add(i);
if((a/i)!=i)
factors.add(a/i);
}
}
System.out.println(factors);
}
public static void main(String args[]){
long x = 24562368L;
Number N = new Number(x);
for(long i : N.factors){
long ProductOfDigits = 1;
long k = i;
while(k!=0){
ProductOfDigits = ProductOfDigits*(k%10);
k = k/10;
}
if(i*ProductOfDigits == N.value)
System.out.println(i);
}
}
}
首先，找出所有因数分解的方法：
100 - 2 * 50 - 4 * 25 - 2 * 2 * 25 - ... 以此类推 ... - 2 * 2 * 5 * 5
如果数字中有任何 1 位，请使用这些添加一些因素：
遍历所有这些因式分解，看看其中是否有正确的形式。
“正确的形式”是一种因式分解，其中一个因子的位数与因子数相同（减去一个），而其他因子等于位数
这提出了一种在找到分解时过滤分解的方法，因为一旦找到
分解是行不通的。
这是一些分解数字的方法的链接：http ://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.117.1230&rep=rep1&type=pdf
这是一些执行此操作的代码。没有关于在所有情况下都正确的承诺。当它无法解决时，使用蛮力分解和一些过滤器来短路过程。
package com.x.factors;
import java.util.ArrayList;
import java.util.List;
public class Factors {
private List<Long> solutions;
private final long original;
private final int originalDigitCount;
private List<Long> primes = new ArrayList<Long>();
public Factors(long original) {
this.original = original;
this.originalDigitCount = ("" + original).length();
}
public List<Long> findSeeds() {
// Consider a number 123, the product of the number with its digits (123*1*2*3 = 738) is 738. Therefore, 123 is
// the seed product of 738. Write a program to accept a number and find all possible seed products. For example,
// If the user entered 4977 then the answer should be 79 and 711.
solutions = new ArrayList<Long>();
// filter out numbers that can't have seeds
// Number must be positive (and not zero)
if (original <= 0) {
return solutions;
}
// Find a number with a 0 digit in it
long temp = original;
while (temp > 0) {
if (temp % 10 == 0) {
return solutions;
}
temp = temp / 10;
}
collectFactors(original, new ArrayList<Long>(), 0, 0);
return solutions;
}
private void collectFactors(long num, List<Long> factors, int factorCount, int doubleDigitFactorCount) {
if (primes.contains(num)) {
return;
}
// The seed can't have more digits than the original number. Thus if we find more factors excluding
// the seed than that, this can't be a solution.
if (factorCount > originalDigitCount) {
return;
}
boolean isPrime = true; // check whether num is prime
int newDoubleDigitFactorCount = 0;
for (long i = num / 2; i > 1; --i) {
// Once we have one factor 2 digits or over, it has to be the seed, so there is no use trying
// any more double digit numbers as only single digits are needed.
if (i > 9) {
if (doubleDigitFactorCount > 0) {
return; // short circuit because of two many non-one-digit factors
}
newDoubleDigitFactorCount = 1;
} else {
newDoubleDigitFactorCount = 0;
}
long remdr = num / i;
if (remdr * i == num) { // is it a factor?
isPrime = false; // it has a factor, its not prime
// add this new factor into the list
if (factors.size() <= factorCount) {
factors.add(i);
} else {
factors.set(factorCount, i);
}
// found a list of factors ... add in the remainder and evaluate
if (factors.size() <= factorCount + 1) {
factors.add(remdr);
} else {
factors.set(factorCount + 1, remdr);
}
long seed = evaluate(factors, factorCount + 2);
if (seed > 0) {
if (solutions.contains(seed)) {
continue;
}
solutions.add(seed);
}
collectFactors(remdr, factors, factorCount + 1, doubleDigitFactorCount + newDoubleDigitFactorCount);
}
}
if (isPrime) { // if its prime, save it
primes.add(num);
}
return;
}
/* package */long evaluate(List<Long> factors, int factorCount) {
// Find seed, the largest factor (or one of them if several are the same)
long seed = 0; // Note seed will be larger than 0
int seedIndex = 0;
for (int i = 0; i < factorCount; ++i) {
if (factors.get(i) > seed) {
seed = factors.get(i);
seedIndex = i;
}
}
// Count the digits in the seed, see if there are the right number of factors. Ignore 1's
boolean[] factorUsed = new boolean[factorCount]; // start off as all false
int seedDigitCount = 0;
long temp = seed;
while (temp > 0) {
if (temp % 10 != 1) {
++seedDigitCount;
}
temp = temp / 10;
}
if (seedDigitCount != factorCount - 1) {
return 0; // fail - seed digit count doesn't equal number of single digit factors
}
// See if all the seed's digits are present
temp = seed;
factorUsed[seedIndex] = true;
while (temp > 0) {
int digit = (int) (temp % 10);
if (digit != 1) { // 1's are never in the factor array, they are just freely ok
boolean digitFound = false;
for (int digitIndex = 0; digitIndex < factorCount; ++digitIndex) {
if (digit == factors.get(digitIndex) && !factorUsed[digitIndex]) {
factorUsed[digitIndex] = true;
digitFound = true;
break;
}
}
if (!digitFound) {
return 0; // fail, a digit in the seed isn't in the other factors
}
}
temp = temp / 10;
}
// At this point we know there are the right number of digits in the seed AND we have
// found all the seed digits in the list of factors
return seed;
}
}
这是一个简单的程序。1秒内获得种子根176852740608。现场演示。
更新：找到 376,352,349 -> 153,642,082,955,760 需要 27 秒。你们呢？我不知道这是否好。
更新 2：@ajb 有一个更快的答案，至少在我做的实验中。但是这个答案的优点是更简单！
#include<iostream>
using namespace std;
typedef long long int Big_Int; // To get a 64-bit int
void root_stem(const Big_Int r, const Big_Int product_of_my_digits, const Big_Int target) {
// There are two rules we can use to prune:
//
// First: The product_of_my_digits must divide into the target.
// If not, return
// Second: The products produced lower in the search true will always be higher
// than those above. Therefore, we should return early if
// my_seed_product is larger than the target
if (target % product_of_my_digits != 0)
return;
Big_Int my_seed_product = r * product_of_my_digits;
if(my_seed_product >= target) {
if (my_seed_product == target) {
cout << r << "\t->\t" << my_seed_product << endl;
}
return;
}
// print all roots, with their products, between 10*r and 10*r + 9
for(Big_Int digit_to_append = 1; digit_to_append<=9; ++digit_to_append) {
root_stem(r*10 + digit_to_append, product_of_my_digits*digit_to_append, target);
}
}
int main() {
root_stem(0,1, 4977);
root_stem(0,1, 24562368);
root_stem(0,1, 176852740608);
return 0;
}
public class Seeds
{
public static void main(String[] args)
{
int num=4977;
if(!seed(num).isEmpty())
System.out.println(seed(num));
else
System.out.println("no seed number");
}
public static List<Integer> seed(int num)
{ List<Integer> list=new ArrayList<Integer>();
for(int i=1; i<=num; i++)
{
if(num%i==0)
{
int factor_digit=1;
int factor = i;
factor_digit=factor_digit*factor;
// when i is the factor, find factors of i and multiply them together to form number
while(factor>=1)
{
factor_digit = factor_digit * (factor%10);
factor = factor/10;
}
if(factor_digit== num)
list.add(i);
}
}
return list;
}
}
*
执行一个程序来确定一个数字是否是另一个数字的种子。如果将 X 乘以它的每个数字等于 Y，则称数字 X 是数字 Y 的种子：123 是 738 的种子，因为 123 1 2*3 = 738 */
class SeedNumber
{
public static void main(String[] args)
{
int seed = 45;
int num = 900;
int seedNum = seed;
int check=1;
for(check=check*seed;seed>0;seed /=10)
{
int rem = seed % 10;
check = check * rem;
}
System.out.println(check);
if(check == num)
System.out.println(seedNum+" is a seed of "+num);
else
System.out.println(seedNum+" is not a seed of "+num);
// Implement your code here
}
}