把2-100的立方都放到一个数组里面去遍历,这样做不会超时.
#include<iostream>
using namespace std;
int main()
{ 	int n,i,j,p,q,a[101];
    cin>>n;
	for(i=2;i<=100;i++)
	{a[i]=i*i*i;
	}//得到2-100的三次放数
	for(i=2;i<=n;i++)
		for(j=2;j<=n;j++)
			for(p=2;p<=n;p++)
				for(q=2;q<=n;q++)
				{	if(a[i]==a[j]+a[p]+a[q] &&a[j]<a[p] &&a[p]<a[q] &&a[q]<a[i])
						cout<<"Cube = "<<i<<", Triple = ("<<j<<","<<p<<","<<q<<")"<<endl;
				}
   return 0;
}