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爱因斯坦曾经提出过这样一道有趣的数学题:
有一个长阶梯,
若每步上2阶,最后剩下1阶;
若每步上3阶,最后剩2阶;
若每步上5阶,最后剩下4阶;
若每步上6阶,最后剩5阶;
只有每步上7阶,最后刚好一阶也不剩。
请问该阶梯至少有多少阶。
来分析一下爱因斯坦的阶梯问题。假设阶梯的个数为minNumber,按照前述的条件,minNumber应该满足如下条件:
minNumber除以2的余数为1;
minNumber除以3的余数为2;
minNumber除以5的余数为4;
minNumber除以6的余数为5;
minNumber除以7的余数为0;
很明显这个数是7的倍数,所以,从7开始,对每个7的倍数进行判断,直到寻找到一个最小的满足条件的数据为止。
package com.joshua317; import java.util.Arrays; public class Main { public static void main(String[] args) { int minNumber; System.out.println("爱因斯阶梯问题"); Jieti jieti = new Jieti(); minNumber = jieti.getMinNum(); System.out.printf("这个阶梯至少有%d阶",minNumber); } } class Jieti { public int getMinNum() { int minNumber = 7; while (true) { if (minNumber%2==1 && minNumber%3==2 && minNumber%5==4 && minNumber%6==5) { break; } minNumber = minNumber+7; } return minNumber; } }
本文为joshua317原创文章,转载请注明:转载自joshua317博客 https://www.joshua317.com/article/79