节俭数字是指其位数严格大于其质数分解中的位数(包括指数)的数字。如果涉及素数因子分解的某个质数的指数为1，则该指数对素数因子分解中的位数不起作用。
节俭数字的一些示例是：
1) 125 = 
, here the number of digits in the number is three (1, 2 and 5) which is strictly greater than the number of digits in its prime factorization which is two (5 and 3).
2) 512 = 
, here the number of digits in the number is three (5, 1 and 2) which is strictly greater than the number of digits in its prime factorization which is two (2 and 9).
3) 1029 = 3 × 
, here the number of digits in the number is four (1, 0, 2 and 9) which is strictly greater than the number of digits its prime factorization which is three (3, 7 and 3).
前几个节俭数字是：125、128、243、256、343、512、625、729等。
这里可能要注意，质数不是节俭数，因为质数的质数分解中的位数等于质数中的位数(因为未考虑值1的指数)。
示例19 = 
，但指数中的1不会影响数字的质数分解中的数字。因此，数字中的位数是两个(1和9)，等于其素数分解中的位数(1和9)。
查找数字n是否节俭的程序涉及简单的步骤。首先，我们找到直到’n’的所有素数，然后找到n的素数分解。最后，我们检查n中的位数是否大于n的素数分解中的位数。
C++
// Program to check for Frugal number
#include
using namespace std;
// Finding primes upto entered number
vector primes(long long int n)
{
bool prime[n + 1];
// Finding primes by Sieve of Eratosthenes method
memset(prime, true, sizeof(prime));
for (int i = 2; i * i <= n; i++) {
// If prime[i] is not changed, then it is prime
if (prime[i] == true) {
// Update all multiples of p
for (int j = i * 2; j <= n; j += i)
prime[j] = false;
}
}
// Forming array of the prime numbers found
vector arr;
for (int i = 2; i < n; i++)
if (prime[i])
arr.push_back(i);
return arr;
}
// Returns number of digits in n
int countDigits(long long int n)
{
long long int temp = n;
int c = 0;
while (temp != 0) {
temp = temp / 10;
c++;
}
return c;
}
// Checking whether a number is Frugal or not
bool frugal(long long int n)
{
vector r = primes(n);
long long int t = n;
// Finding number of digits in prime
// factorization of the number
long long int s = 0;
for (int i = 0; i < r.size(); i++) {
if (t % r[i] == 0) {
// Exponent for current factor
long long int k = 0;
// Counting number of times this prime
// factor divides (Finding exponent)
while (t % r[i] == 0) {
t = t / r[i];
k++;
}
// Finding number of digits in the exponent
// Avoiding exponents of value 1
if (k == 1)
s = s + countDigits(r[i]);
else if (k != 1)
s = s + countDigits(r[i]) + countDigits(k);
}
}
// Checking condition for frugal number
return (countDigits(n) > s && s != 0);
}
// Driver Method to check for frugal number
int main()
{
long long int n = 343;
if (frugal(n))
cout << "A Frugal number\n";
else
cout << "Not a frugal number\n";
return 0;
} Java
// Program to check
// for Frugal number
import java.io.*;
import java.util.*;
class GFG
{
// Finding primes upto
// entered number
static ArrayList
primes(long n)
{
boolean []prime =
new boolean[(int)n + 1];
for(int i = 0;
i < n + 1; i++)
prime[i] = true;
// Finding primes by Sieve
// of Eratosthenes method
for (int i = 2;
i * i <= n; i++)
{
// If prime[i] is not
// changed, then it
// is prime
if (prime[i] == true)
{
// Update all
// multiples of p
for (int j = i * 2;
j <= n; j += i)
prime[j] = false;
}
}
// Forming array of the
// prime numbers found
ArrayList arr =
new ArrayList();
for (int i = 2; i < n; i++)
if (prime[i])
arr.add((long)i);
return arr;
}
// Returns number
// of digits in n
static int countDigits(long n)
{
long temp = n;
int c = 0;
while (temp != 0)
{
temp = temp / 10;
c++;
}
return c;
}
// Checking whether a
// number is Frugal or not
static boolean frugal(long n)
{
ArrayList r = primes(n);
long t = n;
// Finding number of digits
// in prime factorization
// of the number
long s = 0;
for (int i = 0;
i < r.size(); i++)
{
if (t % r.get(i) == 0)
{
// Exponent for
// current factor
long k = 0;
// Counting number of times
// this prime factor divides
// (Finding exponent)
while (t % r.get(i) == 0)
{
t = t / r.get(i);
k++;
}
// Finding number of digits
// in the exponent Avoiding
// exponents of value 1
if (k == 1)
s = s + countDigits(r.get(i));
else if (k != 1)
s = s + countDigits(r.get(i)) +
countDigits(k);
}
}
// Checking condition
// for frugal number
return (countDigits(n) > s && s != 0);
}
// Driver Code
public static void main(String[] args)
{
long n = 343;
if (frugal(n))
System.out.print("A Frugal number\n");
else
System.out.print("Not a frugal number\n");
}
}
// This code is contributed by
// Manish Shaw(manishshaw1) Python3
# Program to check for Frugal number
# Finding primes upto entered number
def primes(n):
# Finding primes by Sieve
# of Eratosthenes method
prime = [True] * (n + 1);
i = 2;
while (i * i <= n):
# If prime[i] is not changed,
# then it is prime
if (prime[i] == True):
# Update all multiples of p
j = i * 2;
while (j <= n):
prime[j] = False;
j += i;
i += 1;
# Forming array of the prime
# numbers found
arr = [];
for i in range(2, n):
if (prime[i]):
arr.append(i);
return arr;
# Returns number of digits in n
def countDigits(n):
temp = n;
c = 0;
while (temp != 0):
temp = int(temp / 10);
c += 1;
return c;
# Checking whether a number is
# Frugal or not
def frugal(n):
r = primes(n);
t = n;
# Finding number of digits
# in prime factorization
# of the number
s = 0;
for i in range(len(r)):
if (t % r[i] == 0):
# Exponent for current factor
k = 0;
# Counting number of times
# this prime factor divides
# (Finding exponent)
while (t % r[i] == 0):
t = int(t / r[i]);
k += 1;
# Finding number of digits
# in the exponent Avoiding
# exponents of value 1
if (k == 1):
s = s + countDigits(r[i]);
elif (k != 1):
s = (s + countDigits(r[i]) +
countDigits(k));
# Checking condition
# for frugal number
return (countDigits(n) > s and s != 0);
# Driver Code
n = 343;
if (frugal(n)):
print("A Frugal number");
else:
print("Not a frugal number");
# This code is contributed by
# mitsC#
// Program to check for Frugal number
using System;
using System.Collections.Generic;
class GFG
{
// Finding primes upto
// entered number
static List primes(long n)
{
bool []prime = new bool[n + 1];
for(int i = 0; i < n + 1; i++)
prime[i] = true;
// Finding primes by Sieve
// of Eratosthenes method
for (int i = 2; i * i <= n; i++)
{
// If prime[i] is not
// changed, then it is prime
if (prime[i] == true)
{
// Update all multiples of p
for (int j = i * 2;
j <= n; j += i)
prime[j] = false;
}
}
// Forming array of the
// prime numbers found
List arr = new List();
for (int i = 2; i < n; i++)
if (prime[i])
arr.Add(i);
return arr;
}
// Returns number of digits in n
static int countDigits(long n)
{
long temp = n;
int c = 0;
while (temp != 0)
{
temp = temp / 10;
c++;
}
return c;
}
// Checking whether a number
// is Frugal or not
static bool frugal(long n)
{
List r = primes(n);
long t = n;
// Finding number of digits in prime
// factorization of the number
long s = 0;
for (int i = 0; i < r.Count; i++)
{
if (t % r[i] == 0)
{
// Exponent for current factor
long k = 0;
// Counting number of times
// this prime factor divides
// (Finding exponent)
while (t % r[i] == 0)
{
t = t / r[i];
k++;
}
// Finding number of digits
// in the exponent Avoiding
// exponents of value 1
if (k == 1)
s = s + countDigits(r[i]);
else if (k != 1)
s = s + countDigits(r[i]) +
countDigits(k);
}
}
// Checking condition
// for frugal number
return (countDigits(n) > s && s != 0);
}
// Driver Code
static void Main()
{
long n = 343;
if (frugal(n))
Console.Write("A Frugal number\n");
else
Console.Write("Not a frugal number\n");
}
}
// This code is contributed by
// Manish Shaw(manishshaw1) PHP
$s &&
$s != 0);
}
// Driver Code
$n = 343;
if (frugal($n))
echo ("A Frugal number\n");
else
echo ("Not a frugal number\n");
// This code is contributed by
// Manish Shaw(manishshaw1)
?>输出：
A Frugal number
优化 ：
上面的代码可以使用“打印所有主要因素及其功效”中讨论的方法进行优化。