ZOJ Problem Set–1331 Perfect Cubes

Time Limit: 10 Seconds      Memory Limit: 32768 KB

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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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Source: Mid-Central USA 1995

#include<iostream>
#include<math.h>
#include<string>
#include<set>
using namespace std;
class Cube
{
public:
  int a, b,c,d;
  Cube(int a, int b,int c, int d):a(a),b(b),c(c),d(d){}
  bool operator<(const Cube& cube) const
  {
    return a<cube.a;
  }
};
const double MINNUMBER = 0.000000001;
int main()
{
  multiset<Cube> s;
  for(int b = 2; b <= 200; b++)
  {
    for(int c = b + 1; c <= 200; c++)
    {
      for(int d = c + 1; d <= 200; d++)
      {
        double ad = pow(b*1.0, 3.0) + pow(c*1.0, 3.0) + pow(d*1.0, 3.0);
        for(int a = d;a <= 200;a++)
        {
          double ad2 = pow(a*1.0, 3);
          if(ad2 > ad + MINNUMBER) break;
          if(ad + MINNUMBER > ad2 && ad - MINNUMBER < ad2)          
          {
            s.insert(Cube(a, b,c,d));
          }
        }
      }
    }
  }
  for(multiset<Cube>::iterator it = s.begin(); it != s.end(); it++)
  {
    cout<<"Cube = "<<it->a<<", Triple = ("<<it->b<<","<<it->c<<","<<it->d<<")"<<endl;
  }
  return 0;
}

看到题目的时间和内存要求，我就笑了，看上去好像没有任何的效率要求，2M的时间，可以容许我干很多事情了。索性也不做那么多事情了，直接用循环解决问题。

不过在这中间还是产生了疑问，本来我是用set的，但是不知道为什么set没有办法把这样的两个元素：Cube（1,2,3), Cube(1,3,4)放进去，这两个元素看上去并不一样啊，让我小小郁闷了一下，改用multiset之后，问题解决了，不过对于set这个容器，还是有上述的疑问，回头有空再研究一下，set到底是凭什么判断两个元素一样的。