原文:

给定一个整数 n，任务是打印 n 的所有子字符串，而不做任何转换，即将其转换为字符串或数组。

示例:

输入 : n = 12345 输出:可能的子串:{1，12，123，1234，12345，2，23，234，2345，3，34，345，4，45，5} 输入 : n = 123 输出:可能的子串:{1，12，123，2，23，3

进场:

  1. 根据大小取 10 的幂。
  2. 将数字除以 0，然后打印出来。
  3. 然后通过取 k 的模，将该数改变到该数的下一个位置。
  4. 更新位数。
  5. 重复同样的过程，直到 n 变成 0。

以下是上述方法的实现:

c 

// c   implementation of above approach
#include 
using namespace std;
// function to print the substrings of a number
void printsubstrings(int n)
{
    // calculate the total number of digits
    int s = log10(n);
    // 0.5 has been added because of it will
    // return double value like 99.556
    int d = (int)(pow(10, s)   0.5);
    int k = d;
    while (n) {
        // print all the numbers from
        // starting position
        while (d) {
            cout << n / d << endl;
            d = d / 10;
        }
        // update the no.
        n = n % k;
        // update the no.of digits
        k = k / 10;
        d = k;
    }
}
// driver code
int main()
{
    int n = 123;
    printsubstrings(n);
    return 0;
}

java 语言(一种计算机语言，尤用于创建网站)

// java implementation
// of above approach
import java.util.*;
import java.lang.*;
import java.io.*;
class gfg
{
// function to print the
// substrings of a number
static void printsubstrings(int n)
{
    // calculate the total
    // number of digits
    int s = (int)math.log10(n);
    // 0.5 has been added because
    // of it will return double
    // value like 99.556
    int d = (int)(math.pow(10, s)   0.5);
    int k = d;
    while (n > 0)
    {
        // print all the numbers
        // from starting position
        while (d > 0)
        {
            system.out.println(n / d);
            d = d / 10;
        }
        // update the no.
        n = n % k;
        // update the no.of digits
        k = k / 10;
        d = k;
    }
}
// driver code
public static void main(string args[])
{
    int n = 123;
    printsubstrings(n);
}
}
// this code is contributed
// by subhadeep

python 3

# python3 implementation of above approach
import math
# function to print the substrings of a number
def printsubstrings(n):
    # calculate the total number of digits
    s = int(math.log10(n));
    # 0.5 has been added because of it will
    # return double value like 99.556
    d = (math.pow(10, s));
    k = d;
    while (n > 0):
        # print all the numbers from
        # starting position
        while (d > 0):
            print(int(n // d));
            d = int(d / 10);
        # update the no.
        n = int(n % k);
        # update the no.of digits
        k = int(k // 10);
        d = k;
# driver code
if __name__ == '__main__':
    n = 123;
    printsubstrings(n);
# this code is contributed by rajput-ji

c

// c# implementation
// of above approach
using system;
class gfg
{
// function to print the
// substrings of a number
static void printsubstrings(int n)
{
    // calculate the total
    // number of digits
    int s = (int)math.log10(n);
    // 0.5 has been added because
    // of it will return double
    // value like 99.556
    int d = (int)(math.pow(10, s)   0.5);
    int k = d;
    while (n > 0)
    {
        // print all the numbers
        // from starting position
        while (d > 0)
        {
            console.writeline(n / d);
            d = d / 10;
        }
        // update the no.
        n = n % k;
        // update the no.of digits
        k = k / 10;
        d = k;
    }
}
// driver code
public static void main()
{
    int n = 123;
    printsubstrings(n);
}
}
// this code is contributed
// by mits

服务器端编程语言（professional hypertext preprocessor 的缩写）

java 描述语言

output: 

1
12
123
2
23
3

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