欧拉计划第6题题解

Sum square difference

The sum of the squares of the first ten natural numbers is,

\[1^2 + 2^2 + \cdots + 10^2 = 385 \]

The square of the sum of the first ten natural numbers is,

\[(1 + 2 + \cdots + 10)^2 = 55^2 = 3025 \]

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

平方的和与和的平方之差

前十个自然数的平方的和是

\[1^2 + 2^2 + \cdots + 10^2 = 385 \]

前十个自然数的和的平方是

\[(1 + 2 + \cdots + 10)^2 = 55^2 = 3025 \]

因此前十个自然数的平方的和与和的平方之差是 3025 − 385 = 2640。

求前一百个自然数的平方的和与和的平方之差。

解题思路

前 \(n\) 个数的和为 \(\frac{n \cdot (n+1)}{2}\)，所以前 \(n\) 个数的和的平方为

\[\frac{n^2 \cdot (n+1)^2}{4} \]

前 \(n\) 个数的平方和公式为

\[\frac{n \cdot (n+1) \cdot (2n+1)}{6} \]

所以本题的答案为

\[\frac{n^2 \cdot (n+1)^2}{4} - \frac{n \cdot (n+1) \cdot (2n+1)}{6} \]

实现代码如下：

#include <bits/stdc++.h>
using namespace std;
long long f(long long n) {
    return n*n*(n+1)*(n+1)/4 - n*(n+1)*(2*n+1)/6;
}
int main() {
    cout << f(100) << endl;
    return 0;
}

答案为 \(25164150\)。