【杭电oj】1334 - Perfect Cubes（水）

Perfect Cubes

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2767 Accepted Submission(s): 1250 Problem Description

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

Output

The output should be listed as shown below, one perfect cube per line, in non-decreasing order 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-decreasing order on the line itself. There do exist several values of a which can be produced from multiple distinct sets of b, c, and d triples. In these cases, the triples with the smaller b values should be listed first. The first part of the output is shown here: 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 efficient implementation. The official time limit for this problem is 2 minutes, and it is indeed possible to write a solution to this problem which executes in under 2 minutes on a 33 MHz 80386 machine. Due to the distributed nature of the contest in this region, judges have been instructed to make the official time limit at their site the greater of 2 minutes or twice the time taken by the judge's solution on the machine being used to judge this problem.

Source

Mid-Central USA 1995

我还以为这样会超时呢，以为有什么数学公式（被学长出的数学题吓怕了）。

用几层for循环直接滚过去就行了，好水的题。

代码如下：

#include <cstdio>
#include <cstring>
int main()
{
	int n = 200;
//	scanf ("%d",&n);
	for (int a=1 ; a<=n ; a++)
	{
		int t = a*a*a;
		for (int b=2 ; b<=n ; b++)
		{
			int t1 = b*b*b;
			if (b >= a)
				break;
			for (int c=b+1 ; c<=n ; c++)
			{
				int t2 = c*c*c;
				if (t1 + t2 >= t)
					break;
				for (int d=c+1 ; d<=n ; d++)
				{
					if (t1 + t2 + d*d*d > t)
						break;
					else if (t1 + t2 + d*d*d == t)
						printf ("Cube = %d, Triple = (%d,%d,%d)\n",a,b,c,d);
				}
			}
		}
	}
	return 0;
}