In-Depth Analysis of GESP Certification C++ Level 1 Questions (Programming Problem — Pyramid) for September 2025

【Problem Description】

The pyramid is constructed withn layers, with each layer requiringn×n,(n−1)×(n−1),…,2×2,1×1 blocks of stone. How many blocks of stone are needed to build the pyramid in total?

【Input】

A single line containing a positive integern, representing the number of layers of the pyramid.

【Output】

A single line containing a positive integer, representing the number of blocks of stone required to build the pyramid.

【Sample Input】

2

【Sample Output】

5

【Hint】

Sample Input 2:

5

Sample Output 2:

55

【Data Range】

For all test cases, it is guaranteed that1≤n≤50.

【Problem Analysis】

This problem requires calculating the total number of stone blocks needed to build an n-layer pyramid. The construction rule is: from the bottom of the pyramid upwards, the i-th layer requires i×i blocks of stone. Therefore, the total number of blocks is equal to the sum of 1² + 2² + 3² + … + n².【Solution Approach】This is a typical problem of calculating the sum of squares. Directly using a loop to accumulate each i² (where i ranges from 1 to n) will yield the result. Since the range of n is from 1 to 50, using a loop is completely feasible and efficient enough.【Algorithm Steps Analysis】

1. Input: Read the number of layers n

2. Initialization: Set accumulator sum=0

3. Loop Accumulation: For i from 1 to n, execute sum += i×i

4. Output: Print the accumulated result sum

【Program Code Analysis】

#include <iostream> using namespace std; typedef long long ll; int main() { ll n, sum = 0; // Define the number of layers n and the accumulated sum sum, using long long to ensure large number calculations cin >> n; // Read the number of layers of the pyramid // Loop to calculate the number of stone blocks needed for each layer and accumulate for(int i = 1; i <= n; i++) { sum += i * i; // The i-th layer requires i² blocks of stone } cout << sum << endl; // Output the total number of blocks return 0; }

Code Detail Explanation:

Variable Type: Use<span><span>long long</span></span> type to store n and sum, ensuring that the calculation result does not overflow when n=50 (50²=2500, total 42925, although both are within int range, it is safer to use long long by convention).

Loop Design: Calculate the number of stone blocks needed for each layer from 1 to n and accumulate, with clear and intuitive logic.

Initialization: Initialize sum to 0 to ensure correct accumulation.

【Optimization Improvements】

Although the current code is efficient enough for n≤50, if the range of n is larger (for example, n≤10⁶), the loop method may time out, and at that point, the sum of squares formula must be used:

sum = n(n+1)(2n+1)/6

This method has a time complexity of O(1) and is highly efficient.