完美立方

 OpenJ_Bailian - 2810 

Time limit1000 ms   　　Memory limit65536 kB　　OS Linux　　Source1543

形如a³= b³ + c³ + d³的等式被称为完美立方等式。例如12³= 6³ + 8³ + 10³ 。编写一个程序，对任给的正整数N (N≤100)，寻找所有的四元组(a, b, c, d)，使得a³ = b³ + c³ + d³，其中a,b,c,d 大于 1, 小于等于N，且b<=c<=d。

Input一个正整数N (N≤100)。Output每行输出一个完美立方。输出格式为： 
Cube = a, Triple = (b,c,d) 
其中a,b,c,d所在位置分别用实际求出四元组值代入。 

请按照a的值，从小到大依次输出。当两个完美立方等式中a的值相同，则b值小的优先输出、仍相同则c值小的优先输出、再相同则d值小的先输出。

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)

 

 1 #include <iostream>
 2 using namespace std;
 3 
 4 //#define cube(a) a * a * a
 5 
 6 
 7 int main()
 8 {
 9     int a, b, c, d, N;
10     cin >> N;
11     int cube[101];
12     for (int i = 2; i < 101; i++)
13         cube[i] = i * i * i;
14 
15     for (a = 2; a <= N; a++)
16     for (b = 2; b < a; b++)
17     for (c = b; c < a; c++)
18     for (d = c; d < a; d++)
19     if (cube[b] + cube[c] + cube[d] == cube[a])
20         cout << "Cube = " << a << ", Triple = (" << b << "," << c << "," << d << ")" << endl;
21 
22     return 0;
23 }