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;
}

  

 

转载于:https://www.cnblogs.com/hxsyl/archive/2012/08/14/2638825.html