计算a^3=b^3+c^3+d^3

Perfect Cubes

Problem Description

For hundredsof years Fermat's Last Theorem, which stated simply that for n > 2 there exist no integers a, b, c > 1 suchthat a^n = b^n + c^n, has remained elusivelyunproven. (A recent proof is believed to be correct, though it is stillundergoing scrutiny.) It is possible, however, to find integers greater than 1that satisfy the ``perfect cube'' equation a^3 = b^3 + c^3 + d^3 (e.g. a quickcalculation will show that the equation 12^3 = 6^3 + 8^3 + 10^3 is indeedtrue). This problem requires that you write a program to find all sets of numbers {a, b, c, d} which satisfythis equation for a <= 200.

Output

The outputshould be listed as shown below, one perfect cube per line, in non-decreasingorder of a (i.e. the lines should be sorted by their a values). The values of b, c, and d should also be listed in non-decreasingorder on the line itself. There do exist several values of a which can beproduced from multiple distinct sets of b, c, and d triples（三方的）. In these cases, thetriples with the smaller b values should belisted first.（由下划线部分可知，求的是a^3=b^3+c^3+d^3,且b最小）

The first part of the output is shown here:（由此句可知，a是从6开始的，所以程序中a最小为6即可）

Cube = 6, Triple = (3,4,5)
Cube = 12, Triple = (6,8,10)
Cube = 18, Triple = (2,12,16)
Cube = 18, Triple = (9,12,15)
Cube = 19, Triple = (3,10,18)
Cube = 20, Triple = (7,14,17)
Cube = 24, Triple = (12,16,20)

Note: The programmer will need to be concerned with an efficientimplementation. The official time limit for this problem is 2 minutes, and itis indeed possible to write a solution to this problem which executes in under2 minutes on a 33 MHz 80386 machine. Due to the distributed nature of thecontest in this region, judges have been instructed to make the official timelimit at their site the greater of 2 minutes or twice the time taken by thejudge's solution on the machine being used to judge this problem

//***********************************
//*  程 序 名：Cube.cpp             *
//*  作    者：何香                   *
//*  编制时间：2013年9月12日        *
//*  主要功能：计算a^3=b^3+c^3+d^3  *
//***********************************

#include <iostream>
using namespace std;
int main()
{
	int a,b,c,d;
	for (a=6;a<=200;++a)
	{
		for (b=2;b<=200;++b)//必须从b开始循环，因为b最小；b<c<d
		{
			for (c=b;c<=200;++c)
			{ 
				for (d=c;d<=200;++d)
				{
					if (a*a*a==b*b*b+c*c*c+d*d*d)
					{
						cout<<"Cube = "<<a<<","<<" Tripe = ("<<b<<","<<c<<","<<d<<")"<<endl;//注意格式
					}
				}
			}
		}
	}
   return 0; 
}