OpenJudge 2810(1543)완벽 한 큐 브/Poj 1543 완벽 한 큐 브

1.링크 주소:
http://bailian.openjudge.cn/practice/2810/
http://bailian.openjudge.cn/practice/1543/
http://poj.org/problem?id=1543
2.제목:
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

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)

Source
Mid-Central USA 1995
3.사고방식:
매 거+타 표(계산 횟수 감소)
a 는 오름차 순 으로 배열 하고 b,c,d 는 오름차 순 으로 배열 해 야 합 니 다.
4.코드:

 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 

37     delete [] arr_cube;

38     return 0;

39 }