4101. Three digit numbers

Find all three-digit numbers with sum of digits equals to n.

Input. One integer n (0 ≤ n ≤ 28).

Output. Print in the first line the number of found three-digit numbers. In the next lines print these numbers in increasing order.

Sample input	Sample output
3	6 102 111 120 201 210 300

SOLUTION

loops

Algorithm analysis

Iterate over the digits of hundreds, tens and ones and find all three-digit numbers, which sum of digits is n.

Algorithm realization

Read the value of n.

scanf("%d",&n);

res = 0;

Iterate through the digits i, j, k of three-digit number and count the number of triples (i, j, k), which sum is n. The number of hundreds i cannot be zero, therefore 1 ≤ i ≤ 9.

for(i = 1; i <= 9; i++)

for(j = 0; j <= 9; j++)

for(k = 0; k <= 9; k++)

if (i + j + k == n) res++;

Print the number of found three-digit numbers.

printf("%d\n",res);

Again iterate over the digits of hundreds, tens and ones and print all three-digit numbers, which sum of digits is n.

for(i = 1; i <= 9; i++)

for(j = 0; j <= 9; j++)

for(k = 0; k <= 9; k++)

if (i + j + k == n) printf("%d%d%d\n", i, j, k);

Java realization

import java.util.*;

public class Main

{

public static void main(String[] args)

{

Scanner con = new Scanner(System.in);

int n = con.nextInt();

int i, j, k, res = 0;

for(i = 1; i <= 9; i++)

for(j = 0; j <= 9; j++)

for(k = 0; k <= 9; k++)

if (i + j + k == n) res++;

System.out.println(res);

for(i = 1; i <= 9; i++)

for(j = 0; j <= 9; j++)

for(k = 0; k <= 9; k++)

if (i + j + k == n) System.out.printf("%d%d%d\n", i, j, k);

con.close();

}

}

Python realization

n = int(input())

res = 0;

for i in range(1,10):

for j in range(0, 10):

for k in range(0, 10):

if (i + j + k == n): res = res + 1;

print(res);

for i in range(1,10):

for j in range(0, 10):

for k in range(0, 10):

if (i + j + k == n): print(str(i) + str(j) + str(k));