Perfect Cubes
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 15302   Accepted: 7936

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.

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)

Java AC 代码

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for(int i = 2; i <= n; i++)
for(int a = 2; a < i; a++)
for(int b = a; b < i; b++)
for(int c = b; c < i; c++) {
if(i*i*i == a*a*a + b*b*b + c*c*c) {
System.out.println("Cube = " + i +", Triple = (" + a + "," + b + "," + c +")");
}
}
}
}

poj 1543 Perfect Cubes (暴搜)的更多相关文章

  1. OpenJudge 2810(1543) 完美立方 / Poj 1543 Perfect Cubes

    1.链接地址: http://bailian.openjudge.cn/practice/2810/ http://bailian.openjudge.cn/practice/1543/ http:/ ...

  2. poj 1543 Perfect Cubes(注意剪枝)

    Perfect Cubes Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14901   Accepted: 7804 De ...

  3. POJ 1543 Perfect Cubes

    Perfect Cubes Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12595   Accepted: 6707 De ...

  4. POJ 1167 The Buses 暴搜+剪枝

    思路: 先把能选的路线都预处理出来 按照能停的车的多少排个序 (剪枝1) 搜搜搜 如果当前剩的车÷当前能停车的多少+deep>=ans剪掉 (剪枝2) //By SiriusRen #inclu ...

  5. POJ 1166 The Clocks (暴搜)

    发现对这样的模拟题根本没啥思路了,本来准备用bfs的.可是结果超时了,这是參考别的人代码写的: #include <stdio.h> #include <iostream> # ...

  6. poj 3080 Blue Jeans(水题 暴搜)

    题目:http://poj.org/problem?id=3080 水题,暴搜 #include <iostream> #include<cstdio> #include< ...

  7. POJ 1945 暴搜+打表 (Or 暴搜+判重)

    思路: 呃呃 暴搜+打表 暴搜的程序::稳稳的TLE+MLE (但是我们可以用来打表) 然后我们就可以打表过了 hiahiahia 可以证明最小的那个数不会超过200(怎么证明的我也不知道),然后就直 ...

  8. [POJ 1204]Word Puzzles(Trie树暴搜&amp;AC自己主动机)

    Description Word puzzles are usually simple and very entertaining for all ages. They are so entertai ...

  9. POJ 1414 暴搜

    题意比较复杂 (但是很好理解) 大概意思是给你等边三角形(详见题目中的图). 最后一行有n个数,下一次要填的数是c. 里面预先已经填好了数字.(0为未填) 得分的标准是这个分数的连通块周围没有空的地方 ...

随机推荐

  1. ANR错误分析

    链接1:https://www.cnblogs.com/xiyuan2016/p/6740623.html 链接2:https://www.jianshu.com/p/3959a601cea6

  2. gcc编译错误

    使用boost的时候遇到一个链接错误 undefined reference to `boost::system::detail::generic_category_instance 出现这个问题的有 ...

  3. 使用KFold进行训练集和验证集的拆分,使用准确率和召回率来挑选合适的阈值(threshold) 1.KFold(进行交叉验证) 2.np.logical_and(两bool数组都是正即为正) 3.np.logical_not(bool数组为正即为反,为反即为正)

    ---恢复内容开始--- 1. k_fold = KFold(n_split, shuffle) 构造KFold的索引切割器 k_fold.split(indices) 对索引进行切割. 参数说明:n ...

  4. MPP - GreenPlum数据库安装以及简单使用

    一.集群介绍 共3台主机,ip 为193.168.0.93   193.168.0.94  193.168.0.95 集群对应master和segment如下,193.168.0.93为master节 ...

  5. 操作MySQL出错提示“BLOB/TEXT column request_data in key specification without a key length”解决办法

    错误原因: 查阅资料后才知道,原来Mysql数据库对于BLOB/TEXT这样类型的数据结构只能索引前N个字符.所以这样的数据类型不能作为主键,也不能是UNIQUE的.所以要换成VARCHAR,但是VA ...

  6. Group By查询

    1.概述 “Group By”从字面意义上理解就是根据“By”指定的规则对数据进行分组,所谓的分组就是将一个“数据集”划分成若干个“小区域”,然后针对若干个“小区域”进行数据处理. 2.原始表 3.简 ...

  7. 慕课网_文件传输基础——Java IO流

    第1章 文件的编码 1-1 文件的编码 (15:07) 第2章 File 类的使用 2-1 File 类常用 API 介绍 (10:50) import java.io.File; import ja ...

  8. 函数对象的apply()和call()方法

    每个函数都包含两个非继承而来的方法:apply()和call().这两个方法的用途都是在特定的作用域中调用函数,特定的作用域为this参数指定的对象. apply()和call()真正强大的地方是能够 ...

  9. 【HTML】---HTML语义化

    1.什么是HTML语义化? <基本上都是围绕着几个主要的标签,像标题(H1~H6).列表(li).强调(strong em)等等> 根据内容的结构化(内容语义化),选择合适的标签(代码语义 ...

  10. video标签在移动端的一些属性值设置

    <video x5-video-orientation="portraint" src="" loop x-webkit-airplay="al ...