求解方程A5+B5+C5+D5+E5=F5

方程A5+B5+C5+D5+E5=F5刚好有一个满足0<A≤B≤C≤D≤E≤F≤75的整数解。请编写一个求出该解的程序：

using System;

namespace ReverseTheExponentiation
{
class Program
{
static void Main(string[] args)
{
Program P = new Program();
P.ReverseTheExponentiation();
}

void ReverseTheExponentiation()
{
int A, B, C, D, E, F;
for (F = 75; F > 0; F--)
{
for (E = 1; E <= F; E++)
{
for (D = 1; D <= E; D++)
{
for (C = 1; C <= D; C++)
{
for (B = 1; B <= C; B++)
{
for (A = 1; A <= B; A++)
{
if (Math.Pow(A, 5) + Math.Pow(B, 5) + Math.Pow(C, 5) + Math.Pow(D, 5) + Math.Pow(E, 5) == Math.Pow(F, 5))
{
Console.WriteLine("A,B,C,D,E,F are: {0},{1},{2},{3},{4},{5}", A, B, C, D, E, F);
}
}
}
}
}
}
}
}
}
}

View Code
输出：



注：如果条件变为0≤A≤B≤C≤D≤E≤F≤75 则有很多解，输出如下：