POJ 1543

//poj1011无限wa，做道简单的,直接暴力竟然AC啦 

//大意：输出所有满足a^3 = b^3 + c^3 + d^3  的a 

#include <iostream>

#include <cstdlib>

using namespace std;

int m(int m)

{

    return m*m*m;

}

int main()

{

    int i,j,k,n,p;

    cin>>n;

    for(p=6;p<=n;p++)

    for(i=2;i<=n-1;i++)//已经保证bcd互不相等 

    for(j=i+1;j<=n-1;j++)

    for(k=j+1;k<=n-1;k++)

    if(m(i)+m(j)+m(k)==m(p))

    cout<<"Cube = "<<p<<","<<" Triple = ("<<i<<","<<j<<","<<k<<")"<<endl;

    //system("pause");

    return 0;

}

  

  

  

//ac

#include <iostream>

#include <cstdlib>

#include <cstdio>

#include <string.h>

using namespace std;

bool vis[101];

int cute[101];

int res[5];

int T,step=1;

void init()

{

    int i,j;

//原来写成了，i = 10，怪不得一直wa
    for(i=1;i<=101;i++)

        cute[i] = i*i*i;

}

void dfs(int num,int start,int n)

{

    int i,j,k;

    if(step==4)

    {

        if(num==0)

            printf("Cube = %d, Triple = (%d,%d,%d)\n",n, res[1], res[2], res[3] );

    }

    else

    {

        for(i=start;cute[i]<=num;i++)//必须有等号，否则无法结束（成立时的最后一个数） 

        if(!vis[i])

        {

            vis[i] = true;

            res[step] = i;

            step++;

            dfs(num-cute[i],i+1,n);

            step--;

            vis[i] = false;

        }

    }

}

    

int main()

{

    int i,j,k;

    init();

    cin>>T;

    memset(vis,false,sizeof(vis));

    for(i=6;i<=T;i++)

    {

        step = 1;

        dfs(cute[i],2,i);

    }

    system("pause");

    return 0;

}