poj 1543 Perfect Cubes

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)

#include <iostream>#include <stdio.h>#include <algorithm>#include <vector>#include <string.h>using namespace std;int n,ans;int a[10];void dfs(int n,int k,int cnt,int now)  //n为当前搜的a值，k为a*a*a减去已经搜到的数后剩余的值，cnt为搜到数的个数，now为前一次搜到的值。{    if(k==0 && cnt==3)    {        printf("Cube = %d, Triple = (%d,%d,%d)\n",n,a[0],a[1],a[2]);        return;    }    if(cnt==3)    {        return;    }    for (int i=now; i<=k; i++)    {        if(k-i*i*i>=0)        {            a[cnt]=i;            dfs(n,k-i*i*i,cnt+1,i);                    }        else        {            return;        }    }}int main(){    while (scanf("%d",&n)!=EOF)    {        for (int i=1; i<=n; i++)        {            dfs(i,i*i*i,0,2);        }    }    return 0;}