2015 Zhegong Da school race -Problem C: Triangle - Fermat's Last Theorem Pythagorean number +

topic

A right triangle has three sides of A, B, C three integers. The longest side length C is known, find a set B, C, B, and C such that the closest.

( Topic Link )

analysis

Hit the table to find the law.

Or a little of the value of direct enumeration of $ CB $. (B unrealistic since enumeration, enumeration difference is a good idea, ah, learned %%% blog )

Or using Fermat's Last Theorem (a small portion),

When ($ A> 1 $) $ A $ is odd, there is an integer of 1 $ $ CB = solutions that $ A ^ 2 + B ^ 2 = C ^ 2 $;

When an even number $ A $ ($ A> 2 $), the presence of $ CB = an integer such that 2 $ solutions $ A ^ 2 + B ^ 2 = C ^ 2 $.

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
ll a, b, c;

int main()
{
    while(scanf("%lld", &a) == 1)
    {
        if(a & 1)
        {
            b = (a * a - 1) / 2;
            c = b + 1;
        }
        else
        {
            b = (a * a - 4) / 4;
            c = b + 2;
        }
        printf("%lld %lld\n", b, c);
    }
    return 0;
}

 

 

Reference links:

1. https://blog.csdn.net/david_jett/article/details/44750687

2. https://blog.csdn.net/Xylon_/article/details/82107186