zoj 1331

Perfect Cubes

Time Limit: 10 Seconds Memory Limit: 32768 KB

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

这道题刚开始看以为有什么公式可以算，结果我还是直接暴力解决问题吧，刚开始用math.h中的pow函数，结果速度超慢的，超时了，最后改了改代码，自己写了一个pow函数，速度虽然比上次的要快，但是还是觉得有点慢，但提交之后居然AC了

#include<stdio.h>
#include<math.h>
int pow(int n1,int n2)
{
if(n2==0)
return 1;
if(n2%2==0)
return pow(n1*n1,n2/2);
else
return pow(n1*n1,n2/2)*n1;
}
int main()
{
int a,b,c,d;
int n=200;
for(a=6;a<=200;a++)
{
for(b=2;b<a;b++)
{
if(pow(b,3)>pow(a,3)/3)
break;
for(c=b;c<a;c++)
{
if(pow(c,3)>(pow(a,3)-pow(b,3))/2)
break;
for(d=c;d<a;d++)
{
if((pow(a,3)-pow(b,3)-pow(c,3))==pow(d,3))
printf("Cube = %d, Triple = (%d,%d,%d)/n", a, b, c, d);
}
}
}
}
return 0;
}