原文:
给定一个数字 n ，打印所有小于或等于 n 的。
强数是一个特殊数，其位数的阶乘之和等于原数。 比如:145 是强号。从，1！ 4! 5!= 145.
示例:
输入: n = 100 输出: 1 2 说明: 只有 1 和 2 是 1 到 100 的强号，因为 1！= 1， 2！= 2
输入: n = 1000 输出: 1 2 145 说明: 只有 1、2、145 是 1 到 1000 的强号，因为 1！= 1， 2！= 2， (1！ 4! 5!) = 145
方法:思路是从【1，n】迭代，检查。如果是，则打印相应的号码，否则检查下一个号码。
下面是上述方法的实现:
c
// c program for the above approach
#include
using namespace std;
// store the factorial of all the
// digits from [0, 9]
int factorial[] = { 1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880 };
// function to return true
// if number is strong or not
bool isstrong(int n)
{
// converting n to string so that
// can easily access all it's digit
string num = to_string(n);
// sum will store summation of
// factorial of all digits
// of a number n
int sum = 0;
for(int i = 0; i < num.length(); i )
{
sum = factorial[num[i] - '0'];
}
// returns true of n is strong number
return sum == n;
}
// function to print all
// strong number till n
void printstrongnumbers(int n)
{
// iterating from 1 to n
for(int i = 1; i <= n; i )
{
// checking if a number is
// strong then print it
if (isstrong(i))
{
cout << i << " ";
}
}
}
// driver code
int main()
{
// given number
int n = 1000;
// function call
printstrongnumbers(n);
return 0;
}
// this code is contributed by rutvik_56
java 语言(一种计算机语言，尤用于创建网站)
// java program for the above approach
class gfg {
// store the factorial of all the
// digits from [0, 9]
static int[] factorial = { 1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880 };
// function to return true
// if number is strong or not
public static boolean isstrong(int n)
{
// converting n to string so that
// can easily access all it's digit
string num = integer.tostring(n);
// sum will store summation of
// factorial of all digits
// of a number n
int sum = 0;
for (int i = 0;
i < num.length(); i ) {
sum = factorial[integer
.parseint(num
.charat(i)
"")];
}
// returns true of n is strong number
return sum == n;
}
// function to print all
// strong number till n
public static void
printstrongnumbers(int n)
{
// iterating from 1 to n
for (int i = 1; i <= n; i ) {
// checking if a number is
// strong then print it
if (isstrong(i)) {
system.out.print(i " ");
}
}
}
// driver code
public static void
main(string[] args)
throws java.lang.exception
{
// given number
int n = 1000;
// function call
printstrongnumbers(n);
}
}
python 3
# python3 program for the
# above approach
# store the factorial of
# all the digits from [0, 9]
factorial = [1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880]
# function to return true
# if number is strong or not
def isstrong(n):
# converting n to string
# so that can easily access
# all it's digit
num = str(n)
# sum will store summation
# of factorial of all
# digits of a number n
sum = 0
for i in range (len(num)):
sum = factorial[ord(num[i]) -
ord('0')]
# returns true of n
# is strong number
if sum == n:
return true
else:
return false
# function to print all
# strong number till n
def printstrongnumbers(n):
# iterating from 1 to n
for i in range (1, n 1):
# checking if a number is
# strong then print it
if (isstrong(i)):
print (i, end = " ")
# driver code
if __name__ == "__main__":
# given number
n = 1000
# function call
printstrongnumbers(n)
# this code is contributed by chitranayal
c
// c# program for the above approach
using system;
class gfg{
// store the factorial of all the
// digits from [0, 9]
static int[] factorial = { 1, 1, 2, 6, 24, 120,
720, 5040, 40320, 362880 };
// function to return true
// if number is strong or not
public static bool isstrong(int n)
{
// converting n to string so that
// can easily access all it's digit
string num = n.tostring();
// sum will store summation of
// factorial of all digits
// of a number n
int sum = 0;
for (int i = 0; i < num.length; i )
{
sum = factorial[int.parse(num[i] "")];
}
// returns true of n is strong number
return sum == n;
}
// function to print all
// strong number till n
public static void printstrongnumbers(int n)
{
// iterating from 1 to n
for (int i = 1; i <= n; i )
{
// checking if a number is
// strong then print it
if (isstrong(i))
{
console.write(i " ");
}
}
}
// driver code
public static void main(string[] args)
{
// given number
int n = 1000;
// function call
printstrongnumbers(n);
}
}
// this code is contributed by sapnasingh4991
java 描述语言
output:
1 2 145
时间复杂度: o(n)
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