OpenJudge 2810(1543) 完美立方 / Poj 1543 Perfect Cubes

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

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.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：

#include <iostream>
#include <cstdio>

#define START_N 2

using namespace std;

int main()
{
    int n;
    cin>>n;

    int *arr_cube = new int[n];

    int i,j,k,p;
    for(i = START_N; i <= n; ++i)
    {
        arr_cube[i - START_N] = i * i * i;
        for(j = START_N; j <= i; ++j)
        {
            for(k = j; k <= i; ++k)
            {
                for(p = k; p <= i; ++p)
                {
                    if(arr_cube[i - START_N] == arr_cube[j - START_N]
                        + arr_cube[k - START_N] + arr_cube[p - START_N])
                    {
                        cout<<"Cube = "<<i<<", Triple = ("<<j<<","<<k<<","<<p<<")"<<endl;
                    }
                }
            }
        }

    }


    delete [] arr_cube;
    return 0;
}