Check Whether a Number Is Strong

PROBLEM UNDERSTANDING

Input and expected output

145

Strong number

COMPLETE C PROGRAM

Complete C implementation

#include <stdio.h>

int main(void)
{
    int number, value, sum = 0;

    printf("Enter a non-negative integer: ");
    scanf("%d", &number);
    if (number < 0) {
        printf("Use a non-negative integer.\n");
        return 0;
    }
    value = number;
    do {
        int digit = value % 10;
        int factorial = 1;
        for (int item = 2; item <= digit; item++) factorial *= item;
        sum += factorial;
        value /= 10;
    } while (value > 0);
    printf(sum == number ? "Strong number\n" : "Not a strong number\n");
    return 0;
}

PROGRAM EXPLANATION

Algorithm and explanation

1. Read a non-negative integer and keep its original value in number.
2. Copy number into value so extracting digits does not destroy the original.
3. Extract a digit with value % 10, calculate its factorial starting from 1, and add it to sum.
4. Remove that digit using value /= 10 and repeat; the do-while also processes input 0.
5. Compare sum with number and print the result.

For 145, the extracted digits are 5, 4 and 1. Their factorials are 120, 24 and 1, so the total is 145 and the number is strong. factorial is reset to 1 for every digit; this also handles 0! = 1 correctly. Input 0 produces a sum of 1 and is not strong. Negative inputs are rejected. With d decimal digits, each digit needs at most eight loop iterations, so the algorithm uses O(d) time and O(1) auxiliary space. This demonstration assumes scanf successfully reads an integer; production input handling should check its return value.

EFFICIENCY

Time and space complexity

DEBUGGING CHECKLIST

Common mistakes

Try it yourself