Problem 39 Integer right triangles （数学）

柔情只为你懂 2022-07-15 02:28 76阅读 0赞

## Integer right triangles ##

### Problem 39 ###

If *p* is the perimeter of a right angle triangle with integral length sides, \{ *a*,*b*,*c*\}, there are exactly three solutions for *p* = 120.

\{20,48,52\}, \{24,45,51\}, \{30,40,50\}

For which value of *p* ≤ 1000, is the number of solutions maximised?

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         <span style="font-size:18px">Answer:</span> 
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       <td><strong><span style="font-size:18px">840</span></strong></td> 
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       <td colspan="2"><span style="color:rgb(153,153,153)"><span style="font-size:18px">Completed on Sat, 29 Oct 2016, 17:04</span></span></td> 
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题解：数学，a+b+c = p，a^2+b^2=c^2 ，化简:p^2 + 2ab = 2p(a+b) ....

代码：

#include<bits/stdc++.h>
    using namespace std;
    int solve(int n)
    {
        int ans = 0;
        for (int a=1;a<n;a++)
        {
            for (int b=a;b<n;b++)
            {
                int c = n-a-b;
                if (c>b)
                {
                	// a+b+c = p, a2+b2=c2 化简:p2 + 2ab = 2p(a+b) 
                    if (n*n+2*a*b == 2*n*(a+b))
                    {
                        ans++;
                    }
                }
            }
        }
        return ans;
    }
     
    int main()
    {
        int max_i=0,max_ans=0;
        for (int i=1;i<=1000;i++)
        {
            int ans = solve(i);
            if( ans > max_ans)
            {
                max_i = i;
                max_ans= ans;
            }
        }
        cout<<max_i<<"  "<<max_ans<<endl;
        return 0;
    }