poj1543Perfect Cubes

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 <= N.

Input

One integer N (N <= 100).

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.

Sample Input

24

Sample Output

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)

代码：

#include <stdio.h>  #include <math.h>  int main()  {      int n,a,b,c,d;      __int64 q[101];      while(scanf("%d",&n)!=EOF)      {          for(int  i=1; i<=n; i++)          {              q[i]=i*i*i;          }          for(a=6; a<=n; a++)              for(b=2; b<a-1; b++)              {                  if(q[a]<q[b]+q[b+1]+q[b+2])                      break;                  for(c=b+1; c<a; c++)                  {                      if(q[a]<q[b]+q[c]+q[c+1])                          break;                      for(d=c+1; d<a; d++)                          if(q[a]==q[b]+q[c]+q[d])                              printf("Cube = %d, Triple = (%d,%d,%d)\n",a,b,c,d);                  }              }      }      return 0;  }  

思路：从小到大，分层求解