1.链接地址:

http://bailian.openjudge.cn/practice/2810/

http://bailian.openjudge.cn/practice/1543/

http://poj.org/problem?id=1543

2.题目:

Perfect Cubes
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 13190   Accepted: 6995

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

  1. 24

Sample Output

  1. Cube = 6, Triple = (3,4,5)
  2. Cube = 12, Triple = (6,8,10)
  3. Cube = 18, Triple = (2,12,16)
  4. Cube = 18, Triple = (9,12,15)
  5. Cube = 19, Triple = (3,10,18)
  6. Cube = 20, Triple = (7,14,17)
  7. Cube = 24, Triple = (12,16,20)

Source

3.思路:

枚举+打表(减少计算次数)

注意a要升序排列,然后b,c,d再升序排列

4.代码:

  1. #include <iostream>
  2. #include <cstdio>
  3.  
  4. #define START_N 2
  5.  
  6. using namespace std;
  7.  
  8. int main()
  9. {
  10. int n;
  11. cin>>n;
  12.  
  13. int *arr_cube = new int[n];
  14.  
  15. int i,j,k,p;
  16. for(i = START_N; i <= n; ++i)
  17. {
  18. arr_cube[i - START_N] = i * i * i;
  19. for(j = START_N; j <= i; ++j)
  20. {
  21. for(k = j; k <= i; ++k)
  22. {
  23. for(p = k; p <= i; ++p)
  24. {
  25. if(arr_cube[i - START_N] == arr_cube[j - START_N]
  26. + arr_cube[k - START_N] + arr_cube[p - START_N])
  27. {
  28. cout<<"Cube = "<<i<<", Triple = ("<<j<<","<<k<<","<<p<<")"<<endl;
  29. }
  30. }
  31. }
  32. }
  33.  
  34. }
  35.  
  36. delete [] arr_cube;
  37. return ;
  38. }

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