如何使用相邻的多个变量实例？求编写a+aa+aaa+aaaa循环程序建议

Hey there! Let's break down how to calculate the sum a + aa + aaa + aaaa using a loop—this is a straightforward problem once you spot the pattern behind building each term.

Core Logic #

Each term in the sequence is built incrementally:

• First term: just a
• Second term: take the first term, multiply by 10, then add a → aa
• Third term: take the second term, multiply by 10, add a → aaa
• Fourth term: repeat the same logic → aaaa

We can use a loop to construct each term step by step and accumulate the total sum as we go.

Python Implementation #

Here's a simple, readable script that does exactly this:

# Get the input number from the user
a = int(input("Enter the number a: "))

total_sum = 0
current_term = 0

# Loop 4 times (once for each term in the sum)
for _ in range(4):
    # Build the current term by shifting the previous term left (multiply by 10) and adding a
    current_term = current_term * 10 + a
    # Add the newly built term to the total
    total_sum += current_term

print(f"The final sum is: {total_sum}")

Test Example

If you input 3, the code calculates 3 + 33 + 333 + 3333 = 3702—the script will output this result correctly.

Java Implementation #

For Java developers, here's an equivalent version:

import java.util.Scanner;

public class RepeatNumberSumCalculator {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Enter the number a: ");
        int a = scanner.nextInt();
        
        int totalSum = 0;
        int currentTerm = 0;
        
        for (int i = 0; i < 4; i++) {
            currentTerm = currentTerm * 10 + a;
            totalSum += currentTerm;
        }
        
        System.out.println("The final sum is: " + totalSum);
        scanner.close();
    }
}

Bonus: Flexible Term Count #

If you ever need to calculate the sum for more or fewer terms (like a + aa + aaaa + aaaaa), just adjust the loop count. For example, replace range(4) in Python with range(n) where n is the number of terms you want—no need to rewrite the whole logic!

This approach keeps your code clean, efficient, and easy to tweak.

内容的提问来源于stack exchange，提问作者Dexidus