[Project Euler] Problem 12

The sequence of triangle numbers is generated by adding the natural numbers. So the 7^th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

 1: 1
 3: 1,3
 6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

寻找第一个因数个数超过500的三角数

计算一个数因数个数的方法，最常见的分解质因数的方法

但编程的时候，寻找质因数是很烦琐的事情

所以我们直接用整除判断该数所有的因数

显然，如果所以小于sqrt(该数)的数中有n个因数，那么这个数的因数有2*n个。

如果该数恰是平方数，那么该数有因数2 *n+1个

   
     

    
      #include 
    
      <
    
      iostream
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;


    
      int
    
       getDivisorNumber(
    
      int
    
       n){
 
    
      int
    
       number
    
      =
    
      0
    
      ;
 
    
      int
    
       divisor 
    
      =
    
       
    
      1
    
      ;
 
    
      int
    
       triangle 
    
      =
    
       n
    
      *
    
      (n
    
      +
    
      1
    
      )
    
      /
    
      2
    
      ;
 
    
      while
    
      (divisor
    
      *
    
      divisor 
    
      <
    
       triangle){
 
    
      if
    
      (triangle
    
      %
    
      divisor 
    
      ==
    
       
    
      0
    
      ){
 number
    
      ++
    
      ;
 }
 divisor
    
      ++
    
      ;
 }
 
 
    
      if
    
      (divisor
    
      *
    
      divisor 
    
      ==
    
       triangle)
 
    
      return
    
       
    
      2
    
      *
    
      number
    
      +
    
      1
    
      ;
 
    
      else
    
       
 
    
      return
    
       
    
      2
    
      *
    
      number;
}


    
      int
    
       main(){
 
    
      int
    
       n
    
      =
    
      1
    
      ;
 
    
      while
    
      (getDivisorNumber(n) 
    
      <=
    
       
    
      500
    
      ){
 n
    
      ++
    
      ;
 }
 cout 
    
      <<
    
       n
    
      *
    
      (n
    
      +
    
      1
    
      )
    
      /
    
      2
    
       
    
      <<
    
       endl;
 
    
      return
    
       
    
      0
    
      ;
}