1543(枚举)

 

完美立方

总时间限制:

1000ms

内存限制:

65536kB

描述

For hundreds of years Fermat's Last Theorem, which stated simply that for n > 2 there exist no integers a, b, c > 1 such that a^n = b^n + c^n, has remained elusively unproven. (A recent proof is believed to be correct, though it is still undergoing scrutiny.) It is possible, however, to find integers greater than 1 that 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 is indeed true). This problem requires that you write a program to find all sets of numbers {a,b,c,d} which satisfy this equation for a <= N.

输入

One integer N (N <= 100).

输出

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

样例输入

24

样例输出

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)

翻译

 问题描述： a3 = b3 + c3 + d3为完美立方等式。例如123 = 63 + 83 + 103 。编写一个程序，对任给的正整数N (N≤100)，寻找所有的四元组(a, b, c, d)，使得a3 = b3 + c3 + d3，其中1<a, b, c, d ≤N。
输入：正整数N (N≤100)
输出：每行输出一个完美立方，按照a的值，从小到大依次输出。当两个完美立方等式中a的值相同，则依次按照b、c、d进行非降升序排列输出，即b值小的先输出、然后c值小的先输出、然后d值小的先输出。

解决方法： G++ 逐一枚举a,b,c,d,

#include<iostream>
using namespace std ;
int main()
{
    int M[101] ;
    int N,i ;
    cin>>N;
    for(i=0;i<=N;i++) M[i] = i*i*i ;

    for(int a=2;a<=N;a++)
    for(int b=2;b<N;b++)
    for(int c=b;c<N;c++)
    for(int d=c;d<N;d++)
     {
       int sum =b*b*b+c*c*c+d*d*d ;
      if(sum==M[a])
     cout<<"Cube = "<<a<<", Triple = ("<<b<<","<<c<<","<<d<<")"<<endl ;
     }
}