题目6：找出前100个自然数的“和的平方”与“平方的和”之差

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题目6：找出前100个自然数的“和的平方”与“平方的和”之差

Sum square difference

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

1² + 2² + ... + 10² = 385

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

(1 + 2 + ... + 10)² = 55² = 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.

题目翻译：

前十个自然数的平方和是：

1² + 2² + ... + 10² = 385

前十个自然数的和的平方是：

(1 + 2 + ... + 10)² = 55² = 3025

所以平方和与和的平方的差是 3025 - 385 = 2640。

那么请找出前 100 个自然数的 “和的平方” 与 “平方的和” 之差？


视频讲解：




思路解析及源码参考（C & Python）：

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zhangjinxuan

#include <bits/stdc++.h>
using namespace std;

long long res = 0;
int n;

int main() {
        //scanf("%d", &n);
        n = 100;
        for (int i = 1; i <= n; ++i) {
                res += i * (i + 1 + n) * (n - i);
        }
        printf("%lld\n", res);
        return 0;
}

Kazimierz

#include <bits/stdc++.h>
using namespace std;

int a,b;

int main()
{
    for(int i=1;i<=100;i++) a += i*i,b += i;
    cout<<b*b-a<<endl;
    return 0;
}

liuhongrun2022

print(sum(range(1, 101)) ** 2 - sum(x ** 2 for x in range(1, 101)))

a8634944

这个简单

auend

good object

猫熊同学

# 答案25164150
sum_of_squares = 0
squares_of_sum = 0

for i in range(1, 101):
    sum_of_squares += pow(i, 2)
    squares_of_sum += i

print(f'前100个自然数的“和的平方”与“平方的和”之差：{pow(squares_of_sum, 2) - sum_of_squares}')

歌者文明清理员

hdpf = sum(range(1, 101)) ** 2
pfdh = sum([i ** 2 for i in range(1, 101)])
print(hdpf - pfdh)

pixie99

0

liangxixin

1

salt_eto

直接模拟

Ulong test6 (Ulong n)
{
    Ulong sumpo2=0;
    Ulong sumpo3=0;
    for(int i=1;i<=n;i++)
    {
      sumpo2+=i*i;
      sumpo3+=i;
    }
    sumpo3*=sumpo3;
    return sumpo3-sumpo2;
}
int main (int argc, char *argv[])
{
  auto res = test6 (100);
  std::cout << res << std::endl;
}

salt_eto

复杂度Ｏ１的数学解

Ulong test6 (Ulong n)
{
    auto sumpow2=[](int n){return (n * (n + 1) * (2*n + 1)) / 6;};
    auto sumpow3=[](int n){return (n*n*(n+1)*(n+1))/4;};
    return sumpow3(n)-sumpow2(n);
}
int main (int argc, char *argv[])
{
  auto res = test6 (100);
  std::cout << res << std::endl;
}

1433391058

test

hejiage

看不明白，先学习下吧，