回溯法Perfect Cubes

本文介绍了一个算法,用于找出所有小于等于给定数值N的整数集合{a, b, c, d}

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Perfect Cubes

Description

For hundreds ofyears Fermat's Last Theorem, which stated simply that for n > 2 there existno integers a, b, c > 1 such that a^n = b^n + c^n, has remained elusivelyunproven. (A recent proof is believed to be correct, though it is stillundergoing scrutiny.) It is possible, however, to find integers greater than 1that 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 isindeed true). This problem requires that you write a program to find all setsof numbers {a,b,c,d} which satisfy this equation for a <= N.

Input

One integer N (N<= 100).

Output

The output shouldbe listed as shown below, one perfect cube per line, in non-decreasing order ofa (i.e. the lines should be sorted by their a values). The values of b, c, andd should also be listed in non-decreasing order on the line itself. There doexist several values of a which can be produced from multiple distinct sets ofb, c, and d triples. In these cases, the triples with the smaller b valuesshould 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)

 

本题很简单,代码如下:

 

#include<iostream>

 

using namespacestd;

 

#define maxN 200

#define maxNN 205

#define maxM 5

 

intvisited[maxNN];

intselected[maxNN];

int ans[maxM];

 

void DFS(int n,intindex);

int main()

{

       int n;

       cin>>n;

       for(int i=6;i<=n;i++)

       {

              for(int j=0;j<=n;j++)

              {

                     visited[j]=0;

                     selected[j]=0;

              }

              DFS(i,0);

       }

       return 0;

}

void DFS(int n,intindex)

{

       if(index==3)

       {

       if((n*n*n==ans[0]*ans[0]*ans[0]+ans[1]*ans[1]*ans[1]+ans[2]*ans[2]*ans[2])&&(selected[ans[0]]*selected[ans[1]]*selected[ans[2]]==0))

              {

                     cout<<"Cube ="<<n<<", Triple =("<<ans[0]<<","<<ans[1]<<","<<ans[2]<<")"<<endl;

                     selected[ans[0]]=selected[ans[1]]=selected[ans[2]]=1;

              }

              return ;

       }

       for(int i=2;i<n;i++)

       {

              if(!visited[i])

              {

                     visited[i]=1;

                     ans[index]=i;

                     DFS(n,index+1);

                     visited[i]=0;

              }

       }

}

 

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