查找a + b + c = 1000的毕达哥拉斯三联体 [英] Find Pythagorean triplet for which a + b + c = 1000

问题描述

毕达哥拉斯三联体是一组三个自然数，一个< b< c，为此， a ² + b ² = c ²

例如3 ² + 4 ² = 9 + 16 = 25 = 5 ² .

确切存在一个毕达哥拉斯三联体，其中a + b + c = 1000. 找到产品abc.

来源: http://projecteuler.net /index.php?section=problems&id=9

我尝试过，但是不知道我的代码出了什么问题.这是我在C中的代码:

#include <math.h>
#include <stdio.h>
#include <conio.h>


void main()
{
    int a=0, b=0, c=0;
    int i;
    for (a = 0; a<=1000; a++)
    {
        for (b = 0; b<=1000; b++)
        {
            for (c = 0; c<=1000; c++)
            {
                if ((a^(2) + b^(2) == c^(2)) && ((a+b+c) ==1000)))
                    printf("a=%d, b=%d, c=%d",a,b,c);
            }
        }
    }
getch();    
}

解决方案

#include <math.h>
#include <stdio.h>

int main()
{
    const int sum = 1000;
    int a;
    for (a = 1; a <= sum/3; a++)
    {
        int b;
        for (b = a + 1; b <= sum/2; b++)
        {
            int c = sum - a - b;
            if ( a*a + b*b == c*c )
               printf("a=%d, b=%d, c=%d\n",a,b,c);
        }
    }
    return 0;
}

说明:

• b = a;
  如果a，b(a< = b)和c是毕达哥拉斯三联体，
  然后b，a(b> = a)和c-也是解，所以我们只能搜索一种情况
• c = 1000-a-b; 这是问题的条件之一(我们不需要扫描所有可能的"c":只需对其进行计算)

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a² + b² = c²

For example, 3² + 4² = 9 + 16 = 25 = 5².

There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.

Source: http://projecteuler.net/index.php?section=problems&id=9

I tried but didn't know where my code went wrong. Here's my code in C:

#include <math.h>
#include <stdio.h>
#include <conio.h>


void main()
{
    int a=0, b=0, c=0;
    int i;
    for (a = 0; a<=1000; a++)
    {
        for (b = 0; b<=1000; b++)
        {
            for (c = 0; c<=1000; c++)
            {
                if ((a^(2) + b^(2) == c^(2)) && ((a+b+c) ==1000)))
                    printf("a=%d, b=%d, c=%d",a,b,c);
            }
        }
    }
getch();    
}

解决方案

#include <math.h>
#include <stdio.h>

int main()
{
    const int sum = 1000;
    int a;
    for (a = 1; a <= sum/3; a++)
    {
        int b;
        for (b = a + 1; b <= sum/2; b++)
        {
            int c = sum - a - b;
            if ( a*a + b*b == c*c )
               printf("a=%d, b=%d, c=%d\n",a,b,c);
        }
    }
    return 0;
}

explanation:

• b = a;
  if a, b (a <= b) and c are the Pythagorean triplet,
  then b, a (b >= a) and c - also the solution, so we can search only one case
• c = 1000 - a - b; It's one of the conditions of the problem (we don't need to scan all possible 'c': just calculate it)

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