许多N被称为浪费数。如果数字在N(包括权力)的因式分解的数量超过以N位数。所有素数都是浪费的。
例如，52是一个浪费数字，因为它的质数分解为2 * 2 * 13 ，并且其质数分解的总数为四位数(1、2、2、3)，大于52中的位数。
前几个浪费数字是：
4, 6, 8, 9, 12, 18, 20, 22, 24,…
程序打印小于或等于给定数字N的浪费数字
给定数字N ，任务是打印小于或等于N的浪费数字。
例子：
Input: N = 10
Output: 4, 6, 8, 9
Input: N = 12
Output: 4, 6, 8, 9, 12
方法：
- 使用Sundaram筛网计算所有的质数，直到10 6 。
- 找出N中的位数。
- 找到N的所有素数，并对每个素数p执行以下操作：
- 找出p中的位数。
- 计算除以N的p的最高幂。
- 找到以上两个的总和。
- 如果素数中的位数大于原始数中的位数，则返回true。否则返回false。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
const int MAX = 10000;
// Array to store all prime less
// than and equal to MAX.
vector primes;
// Function for Sieve of Sundaram
void sieveSundaram()
{
// Boolean Array
bool marked[MAX / 2 + 1] = { 0 };
// Mark all numbers which do not
// generate prime number by 2*i+1
for (int i = 1;
i <= (sqrt(MAX) - 1) / 2; i++) {
for (int j = (i * (i + 1)) << 1;
j <= MAX / 2;
j = j + 2 * i + 1) {
marked[j] = true;
}
}
// Since 2 is a prime number
primes.push_back(2);
// Print remaining primes are of the
// form 2*i + 1 such that marked[i]
// is false.
for (int i = 1; i <= MAX / 2; i++)
if (marked[i] == false)
primes.push_back(2 * i + 1);
}
// Function that returns true if n
// is a Wasteful number
bool isWasteful(int n)
{
if (n == 1)
return false;
// Count digits in original number
int original_no = n;
int sumDigits = 0;
while (original_no > 0) {
sumDigits++;
original_no = original_no / 10;
}
int pDigit = 0, count_exp = 0, p;
// Count all digits in prime factors of N
// pDigit is going to hold this value.
for (int i = 0; primes[i] <= n / 2; i++) {
// Count powers of p in n
while (n % primes[i] == 0) {
// If primes[i] is a prime factor,
p = primes[i];
n = n / p;
// Count the power of prime factors
count_exp++;
}
// Add its digits to pDigit
while (p > 0) {
pDigit++;
p = p / 10;
}
// Add digits of power of
// prime factors to pDigit.
while (count_exp > 1) {
pDigit++;
count_exp = count_exp / 10;
}
}
// If n!=1 then one prime factor
// still to be summed up
if (n != 1) {
while (n > 0) {
pDigit++;
n = n / 10;
}
}
// If digits in prime factors is more than
// digits in original number then
// return true. Else return false.
return (pDigit > sumDigits);
}
// Function to print Wasteful Number before N
void Solve(int N)
{
// Iterate till N and check if i
// is wastefull or not
for (int i = 1; i < N; i++) {
if (isWasteful(i)) {
cout << i << " ";
}
}
}
// Driver code
int main()
{
// Precompute prime numbers upto 10^6
sieveSundaram();
int N = 10;
// Function Call
Solve(N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
static int MAX = 10000;
// Array to store all prime less
// than and equal to MAX.
static Vector primes = new Vector();
// Function for Sieve of Sundaram
static void sieveSundaram()
{
// Boolean Array
boolean marked[] = new boolean[MAX / 2 + 1];
// Mark all numbers which do not
// generate prime number by 2*i+1
for (int i = 1;
i <= (Math.sqrt(MAX) - 1) / 2; i++)
{
for (int j = (i * (i + 1)) << 1;
j <= MAX / 2;
j = j + 2 * i + 1)
{
marked[j] = true;
}
}
// Since 2 is a prime number
primes.add(2);
// Print remaining primes are of the
// form 2*i + 1 such that marked[i]
// is false.
for (int i = 1; i <= MAX / 2; i++)
if (marked[i] == false)
primes.add(2 * i + 1);
}
// Function that returns true if n
// is a Wasteful number
static boolean isWasteful(int n)
{
if (n == 1)
return false;
// Count digits in original number
int original_no = n;
int sumDigits = 0;
while (original_no > 0)
{
sumDigits++;
original_no = original_no / 10;
}
int pDigit = 0, count_exp = 0, p = 0;
// Count all digits in prime factors of N
// pDigit is going to hold this value.
for (int i = 0; primes.get(i) <= n / 2; i++)
{
// Count powers of p in n
while (n % primes.get(i) == 0)
{
// If primes[i] is a prime factor,
p = primes.get(i);
n = n / p;
// Count the power of prime factors
count_exp++;
}
// Add its digits to pDigit
while (p > 0)
{
pDigit++;
p = p / 10;
}
// Add digits of power of
// prime factors to pDigit.
while (count_exp > 1)
{
pDigit++;
count_exp = count_exp / 10;
}
}
// If n!=1 then one prime factor
// still to be summed up
if (n != 1)
{
while (n > 0)
{
pDigit++;
n = n / 10;
}
}
// If digits in prime factors is more than
// digits in original number then
// return true. Else return false.
return (pDigit > sumDigits);
}
// Function to print Wasteful Number before N
static void Solve(int N)
{
// Iterate till N and check if i
// is wastefull or not
for (int i = 1; i < N; i++)
{
if (isWasteful(i))
{
System.out.print(i + " ");
}
}
}
// Driver code
public static void main(String[] args)
{
// Precompute prime numbers upto 10^6
sieveSundaram();
int N = 10;
// Function Call
Solve(N);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program for the
# above approach
import math
MAX = 10000
# Array to store all prime
# less than and equal to MAX.
primes = []
# Function for Sieve of
# Sundaram
def sieveSundaram():
# Boolean Array
marked = [False] * ((MAX // 2) + 1)
# Mark all numbers which do not
# generate prime number by 2*i+1
for i in range(1, ((int(math.sqrt(MAX)) -
1) // 2) + 1):
j = (i * (i + 1)) << 1
while j <= (MAX // 2):
marked[j] = True
j = j + 2 * i + 1
# Since 2 is a prime number
primes.append(2)
# Print remaining primes are of the
# form 2*i + 1 such that marked[i]
# is false.
for i in range(1, (MAX // 2) + 1):
if marked[i] == False:
primes.append(2 * i + 1)
# Function that returns true if n
# is a Wasteful number
def isWasteful(n):
if (n == 1):
return False
# Count digits in original
# number
original_no = n
sumDigits = 0
while (original_no > 0):
sumDigits += 1
original_no = original_no // 10
pDigit, count_exp, p = 0, 0, 0
# Count all digits in prime
# factors of N pDigit is going
# to hold this value.
i = 0
while(primes[i] <= (n//2)):
# Count powers of p in n
while (n % primes[i] == 0):
# If primes[i] is a prime
# factor,
p = primes[i]
n = n // p
# Count the power of prime
# factors
count_exp += 1
# Add its digits to pDigit
while (p > 0):
pDigit += 1
p = p // 10
# Add digits of power of
# prime factors to pDigit.
while (count_exp > 1):
pDigit += 1
count_exp = count_exp // 10
i += 1
# If n!=1 then one prime factor
# still to be summed up
if (n != 1):
while (n > 0):
pDigit += 1
n = n // 10
# If digits in prime factors
# is more than digits in
# original number then return
# true. Else return false.
return bool(pDigit > sumDigits)
# Function to print Wasteful Number
# before N
def Solve(N):
# Iterate till N and check
# if i is wastefull or not
for i in range(1, N):
if (isWasteful(i)):
print(i, end = " ")
# Driver code
# Precompute prime numbers
# upto 10^6
sieveSundaram()
N = 10
# Function Call
Solve(N)
# This code is contributed by divyeshrabadiya07C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
static int MAX = 10000;
// Array to store all prime less
// than and equal to MAX.
static List primes = new List();
// Function for Sieve of Sundaram
static void sieveSundaram()
{
// Boolean Array
bool []marked = new bool[MAX / 2 + 1];
// Mark all numbers which do not
// generate prime number by 2*i+1
for (int i = 1;
i <= (Math.Sqrt(MAX) - 1) / 2; i++)
{
for (int j = (i * (i + 1)) << 1;
j <= MAX / 2;
j = j + 2 * i + 1)
{
marked[j] = true;
}
}
// Since 2 is a prime number
primes.Add(2);
// Print remaining primes are of the
// form 2*i + 1 such that marked[i]
// is false.
for (int i = 1; i <= MAX / 2; i++)
if (marked[i] == false)
primes.Add(2 * i + 1);
}
// Function that returns true if n
// is a Wasteful number
static bool isWasteful(int n)
{
if (n == 1)
return false;
// Count digits in original number
int original_no = n;
int sumDigits = 0;
while (original_no > 0)
{
sumDigits++;
original_no = original_no / 10;
}
int pDigit = 0, count_exp = 0, p = 0;
// Count all digits in prime factors of N
// pDigit is going to hold this value.
for (int i = 0; primes[i] <= n / 2; i++)
{
// Count powers of p in n
while (n % primes[i] == 0)
{
// If primes[i] is a prime factor,
p = primes[i];
n = n / p;
// Count the power of prime factors
count_exp++;
}
// Add its digits to pDigit
while (p > 0)
{
pDigit++;
p = p / 10;
}
// Add digits of power of
// prime factors to pDigit.
while (count_exp > 1)
{
pDigit++;
count_exp = count_exp / 10;
}
}
// If n!=1 then one prime factor
// still to be summed up
if (n != 1)
{
while (n > 0)
{
pDigit++;
n = n / 10;
}
}
// If digits in prime factors is more than
// digits in original number then
// return true. Else return false.
return (pDigit > sumDigits);
}
// Function to print Wasteful Number before N
static void Solve(int N)
{
// Iterate till N and check if i
// is wastefull or not
for (int i = 1; i < N; i++)
{
if (isWasteful(i))
{
Console.Write(i + " ");
}
}
}
// Driver code
public static void Main(String[] args)
{
// Precompute prime numbers upto 10^6
sieveSundaram();
int N = 10;
// Function Call
Solve(N);
}
}
// This code is contributed by 29AjayKumar 输出：
4 6 8 9
时间复杂度： O(N * log(log N))
参考： https : //oeis.org/A046760