Fzu1669 right angled triangle

Topic link ：

​ ​http://acm.fzu.edu.cn/problem.php?pid=1669​​

The main idea of the topic ：

Be content to a、b It's a right angle side ,c It's a hypotenuse , And satisfy a + b + c <= L The number of right triangle .

Ideas ：

Pythagorean theorem ,a、b、c That is, the primitive Pythagorean triple , Then meet ：

x = m^2 - n^2

y = 2*m*n

z = m^2 + n^2

among m > n, And if m It's odd , be n For the even , if m For the even , be n It's odd .

enumeration m、n, Then multiply the triple by i times , Guarantee i * (x + y + z) Within the scope given (2 * m^2 + 2 * m*n <= L),

You can find all triples that meet the conditions .

AC Code ：

Log in and copy

      
      

       
       #include<iostream>
       
       

       
       #include<algorithm>
       
       

       
       #include<cstdio>
       
       

       
       #include<cstring>
       
       

       
       #include<cmath>
       
       

       
       using 
       
       namespace 
       
       std;
       
       

       
       

       
       bool 
       
       flag[
       
       1001000];
       
       

       
       

       
       int 
       
       GCD(
       
       int 
       
       a,
       
       int 
       
       b)
       
       
{
       
       
    
       
       if(
       
       b 
       
       == 
       
       0)
       
       
        
       
       return 
       
       a;
       
       
    
       
       return 
       
       GCD(
       
       b,
       
       a
       
       %
       
       b);
       
       
}
       
       

       
       

       
       int 
       
       main()
       
       
{
       
       
    
       
       int 
       
       N;
       
       
    
       
       while(
       
       cin 
       
       >> 
       
       N)
       
       
    {
       
       
        
       
       int 
       
       temp,
       
       m,
       
       n,
       
       i,
       
       ans,
       
       x,
       
       y,
       
       z;
       
       
        
       
       ans 
       
       = 
       
       0;
       
       
        
       
       memset(
       
       flag,
       
       false,
       
       sizeof(
       
       flag));
       
       
        
       
       temp 
       
       = 
       
       sqrt(
       
       N
       
       *
       
       1.0);
       
       
        
       
       for(
       
       n 
       
       = 
       
       1; 
       
       n 
       
       <= 
       
       temp; 
       
       ++
       
       n)
       
       
        {
       
       
            
       
       for(
       
       m 
       
       = 
       
       n
       
       +
       
       1; 
       
       m 
       
       <= 
       
       temp; 
       
       ++
       
       m)
       
       
            {
       
       
                
       
       if(
       
       2
       
       *
       
       m
       
       *
       
       m 
       
       + 
       
       2
       
       *
       
       m
       
       *
       
       n 
       
       > 
       
       N)
       
       
                    
       
       break;
       
       
                
       
       if((
       
       n
       
       &
       
       1) 
       
       != (
       
       m
       
       &
       
       1))
       
       
                {
       
       
                    
       
       if(
       
       GCD(
       
       m,
       
       n) 
       
       == 
       
       1)
       
       
                    {
       
       
                        
       
       x 
       
       = 
       
       m
       
       *
       
       m 
       
       - 
       
       n
       
       *
       
       n;
       
       
                        
       
       y 
       
       = 
       
       2
       
       *
       
       m
       
       *
       
       n;
       
       
                        
       
       z 
       
       = 
       
       m
       
       *
       
       m 
       
       + 
       
       n
       
       *
       
       n;
       
       
                        
       
       for(
       
       int 
       
       i 
       
       = 
       
       1; ; 
       
       ++
       
       i)
       
       
                        {
       
       
                            
       
       if(
       
       i
       
       *(
       
       x
       
       +
       
       y
       
       +
       
       z) 
       
       > 
       
       N)
       
       
                                
       
       break;
       
       
                            
       
       ans
       
       ++;
       
       
                        }
       
       

       
       
                    }
       
       
                }
       
       
            }
       
       
        }
       
       
        
       
       cout 
       
       << 
       
       ans 
       
       << 
       
       endl;
       
       
    }
       
       

       
       
    
       
       return 
       
       0;
       
       
}
      
      
      
      

       
       
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