Problem 6 Sum square difference （数学）

Sum square difference

Problem 6

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

1 ² + 2 ² + ... + 10 ² = 385

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

(1 + 2 + ... + 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.

题解：sum1(n) = n(n + 1)/2.

           sum2(n)=n/6 *(2n + 1)(n + 1).

代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
  ll a=0;
  a=100*(100+1)/2;
  cout<<a*a<<endl;
  ll b=0;
  for(int i=1;i<=100;i++)
  {
    b+=i*i;
  }
  cout<<b<<endl;
  cout<<"ans="<<a*a-b<<endl;
  return 0;
}