[Project Euler] 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.

好吧，我们用最傻瓜的方式来做这道题

   
     

    
      #include 
    
      <
    
      iostream
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;


    
      int
    
       main(){
 
    
      int
    
       sumOfTheSquare 
    
      =
    
       
    
      0
    
      ;
 
    
      int
    
       squareOfTheSum 
    
      =
    
       
    
      0
    
      ;
 
    
      int
    
       sum 
    
      =
    
       
    
      0
    
      ;
 
 
    
      for
    
       (
    
      int
    
       i
    
      =
    
      1
    
      ;i
    
      <=
    
      100
    
      ;i
    
      ++
    
      ){
 sumOfTheSquare 
    
      +=
    
      i
    
      *
    
      i;
 sum 
    
      +=
    
       i;
 }
 squareOfTheSum 
    
      =
    
       sum
    
      *
    
      sum;
 cout 
    
      <<
    
       squareOfTheSum
    
      -
    
      sumOfTheSquare 
    
      <<
    
       endl;
 
    
      return
    
       
    
      0
    
      ;
}
   
     

好吧，你小时候一定不厌其烦的听你的老师给你讲高斯做加法的故事了

不管是真是假，我们有高斯公式和平方和公式

更简洁的公式法是

   
     

    
      #include 
    
      <
    
      iostream
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;


    
      int
    
       getSumOfTheSquare(
    
      int
    
       n){
 
    
      return
    
       n
    
      *
    
      (n
    
      +
    
      1
    
      )
    
      *
    
      (
    
      2
    
      *
    
      n
    
      +
    
      1
    
      )
    
      /
    
      6
    
      ;
}


    
      int
    
       getSquareOfTheSum(
    
      int
    
       n){
 
    
      int
    
       sum 
    
      =
    
       n
    
      *
    
      (n
    
      +
    
      1
    
      )
    
      /
    
      2
    
      ;
 
    
      return
    
       sum
    
      *
    
      sum;
}


    
      int
    
       main(){
 cout 
    
      <<
    
       getSquareOfTheSum(
    
      100
    
      )
    
      -
    
      getSumOfTheSquare(
    
      100
    
      ) 
    
      <<
    
       endl;
 
    
      return
    
       
    
      0
    
      ;
}
   
     

好吧，你也许会说，我可以直接把“和的平方”与“平方之和”的差用公式表示出来，可以一步求出结果

那就随便吧。