joj1611

 1611: Perfect Cubes

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	763	314	Standard

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  , 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  (e.g. a quick calculation will show that the equation  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  .

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.

这个题就是浮云了，不多说了

#include<stdio.h>
void out(int a,int b,int c,int d)
{
    printf("Cube = %d, Triple = (%d,%d,%d)\n",a,b,c,d);
}
bool exam(int a,int b,int c,int d)
{
    if(a*a*a==(b*b*b+c*c*c+d*d*d))
    return true;
    else
    return false;
}
int main()
{
    for(int i=4;i<=200;i++)
    {
        for(int j=2;j<i;j++)
        {
            for(int u=j+1;u<i;u++)
            {
                for(int v=u+1;v<i;v++)
                {
                    if(exam(i,j,u,v))
                    out(i,j,u,v);
                }
            }
        }
    }
    return 0;
}