题目链接：http://poj.org/problem?id=1543
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
题目链接：
寻找：a^3 = b^3 + c^3 + d^3 。
代码如下：（POJ）
#include <cstdio>
#include <cstring>
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i = 3; i <= n; i++)
{
for(int j = 2; j < n; j++)
{
for(int k = j+1; k < n; k++)
{
for(int l = k+1; l < n; l++)
{
if(i*i*i==j*j*j+k*k*k+l*l*l)
printf("Cube = %d, Triple = (%d,%d,%d)\n",i,j,k,l);
}
}
}
}
}
return 0;
}
（HDU & ZOJ）
#include <cstdio>
#include <cstring>
int main()
{
for(int i = 3; i <= 200; i++)
{
for(int j = 2; j < 200; j++)
{
for(int k = j+1; k < 200; k++)
{
for(int l = k+1; l < 200; l++)
{
if(i*i*i==j*j*j+k*k*k+l*l*l)
printf("Cube = %d, Triple = (%d,%d,%d)\n",i,j,k,l);
}
}
}
}
}