ZOJ 1331 Perfect Cubes

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.

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注意回溯的时候减到当前的项。

 

Code
#include<stdio.h>
#define cube(a) (a)*(a)*(a)

int main()
{
    long a, b, c, d, now, sum;

    for (a=1; a<201; a++)
    {
        sum = cube(a);
        b = 2; now = cube(2);
        while (now<sum)
        {
            c = b;
            now += cube(c);
            while (now<sum)
            {
                d = c;
                now += cube(d);
                while (now<sum)
                {
                    d++; now+= cube(d) - cube(d-1);
                }
                if (now==sum) printf("Cube = %d, Triple = (%d,%d,%d)\n", a, b, c, d);
                now -= cube(d);
                c++; now+= cube(c) - cube(c-1);
            }
            now -= cube(c);
            b++; now+= cube(b) - cube(b-1);
        }
    }

    return 0;
}

 

转载于:https://www.cnblogs.com/menie/archive/2009/06/12/1501920.html