Problem 39 Integer right triangles （数学）

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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?

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;}

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