//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 
     
      #include 
      
       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 = "<
       
<<","<<" Triple = ("<<<","<
         
          <<","<
          
           <<")"<
           
            #include 
            
             #include 
             
              #include 
              
               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;}