Perfect Cubes
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 16522 | Accepted: 8444 |
Description
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.
Input
One integer N (N <= 100).
Output
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.
描述
数百年来费马的最后定理，其简单地说明，对于n> 2，没有整数a，b，c> 1使得a ^ n = b ^ n + c ^ n仍然难以确认。（最近的证据被认为是正确的，尽管它仍在进行详细审查。）然而，有可能找到满足“完美立方”方程的大于1的整数a ^ 3 = b ^ 3 + c ^ 3 + d ^ 3（例如，快速计算将显示等式12 ^ 3 = 6 ^ 3 + 8 ^ 3 + 10 ^ 3确实为真）。这个问题要求你编写一个程序来查找满足<= N的这个等式的所有数字{a，b，c，d}。
输入
一个整数N（N <= 100）。
产量
输出应如下所示列出，每行一个完美的立方体，以a的非递减顺序排列（即行应按其值排序）。b，c和d的值也应该在线本身上以非递减顺序列出。确实存在几个可以从多个不同的b，c和d三元组产生的a值。在这些情况下，应首先列出b值较小的三元组。
Sample Input
24
Sample Output
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)
Source
#include <stdio.h>
int main(){
int n,a,b,c,d;
scanf("%d",&n);
for(a=2;a<=n;++a)
for(b=2;b<a;++b)
for(c=b;c<a;++c)
for(d=c;d<a;++d)
if(a*a*a==b*b*b+c*c*c+d*d*d)
printf("Cube = %d, Triple = (%d,%d,%d)\n",a,b,c,d);
return 0;
}