欧拉工程第43题：Sub-string divisibility

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题目链接：https://projecteuler.net/problem=43

0到9的排列所有形成的数中，求满足下来条件的数的和
Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:

d2d3d4=406 is divisible by 2
d3d4d5=063 is divisible by 3
d4d5d6=635 is divisible by 5
d5d6d7=357 is divisible by 7
d6d7d8=572 is divisible by 11
d7d8d9=728 is divisible by 13
d8d9d10=289 is divisible by 17

Java写的，代码戳的一比，开始还是输出所有结果，然后，然后用R算的，后来搞个全局变量，这样就可以求和，但是输出来的有中间的和，最后一个才符合要求。

求排列的方法整理网上的，内涵还是没有理解。

Java代码：

package projecteuler41to50;import java.util.Date;import java.util.Set;import java.util.TreeSet;class OOOO {      public static long sum=0;    }class level43{    void solve(){        String[] numStr={"1","2","3","4","5","6","7","8","9","0"};        permutation(numStr,0,9,0);    }    boolean divisible(String[] numStr,int[] div){        for(int i=1;i<=7;i++){            int d=Integer.valueOf(numStr[i]+numStr[i+1]+numStr[i+2]);            if(d%div[i-1]!=0){                return false;            }        }        return true;    }    void permutation(String[] str,int start,int end,int sum1){        int[] div={2,3,5,7,11,13,17};        if(start==end){             if(divisible(str,div)){                 StringBuffer sb=new StringBuffer();                 for(int i=0;i<=9;i++)                     sb.append(str[i]);                 String res=sb.toString();//               Set set=new TreeSet();//               set.add(res);                 OOOO.sum+=Long.valueOf(res);                 System.out.println(OOOO.sum+" "+res);             }        }        for(int i=start;i<=end;i++){//交换数组第一个元素与后续的元素              String temp=str[start];            str[start]=str[i];            str[i]=temp;            permutation(str,start+1,end,0);//后续元素递归全排列              temp=str[start];//将交换后的数组还原              str[start]=str[i];            str[i]=temp;        }    }}public class Problem43 {    public static void main(String[] args){        Date beginTime=new Date();        new level43().solve();        Date endTime=new Date();        Long Time=endTime.getTime()-beginTime.getTime();        System.out.println("Time="+Time/1000+"秒"+Time%1000+"毫秒");    }}

结果：

1430952867 14309528672891310156 14603572894297667445 14063572898428620312 413095286712588977601 416035728916695334890 4106357289Time=2秒179毫秒

仿照Java代码，写的Python代码
Python 有个好处，就是可以直接导入模块来求出排列

Python代码：

def divis(num):    primes=[2,3,5,7,11,13,17]    for i in range(1,8):        p=str(num)        digit=int(p[i:(i+3)])         if digit%primes[i-1]!=0:            return False    return Truefrom itertools import permutationsfrom time import time t1=time()digits=[ int(''.join(p))  for p in itertools.permutations("0123456789") ]result=0for x in digits:    if divis(int(x)):        result+=int(x)t2=time()   elapsed = t2 - t1print("Found " + str(result)+" in " + str(round(elapsed, 5)) + " seconds")

结果：

Found 16695334890 in 24.038 seconds

24s。。。。。什么也不说了

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