Problem32

We shall say that an n-digit number is pandigital if it makes use of  all the digits 1 to n exactly once; for example, the 5-digit number,  15234, is 1 through 5 pandigital.

 The product 7254 is unusual, as the identity, 39*186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

 Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way
so be sure to only include it once in your sum.

* 根据题意分析：
* 因为A*B=C，而且A、B不能等于1，所以C>B,C>A。
* A的位数+B的位数+C的位数=9，所以C大于3位，C>=1000。
* A的位数+B的位数=9-C的位数，若C为4位，则有两种组合(A为1位，B为4位；A为2位，B为3位)可以满足。
* 若C为5位，则A的位数+B的位数=4，不可能满足A*B=C的条件。
*
* 所以C的范围为（1000-9999）
* A/B的范围为(2-4999),因为A、B位置可互换，
* 所以A的取值范围可确定(2-99),B的取值范围(100-4999)

 

 

import java.util.ArrayList;import java.util.List;public class Problem32 {//判断这三个数能否组成1-9的全数码。利用list来比较，list初始化了1-9的数字。static boolean isPandigital(int a,int b,int c){boolean flag = false;List<Integer> list = new ArrayList<Integer>();for(int i=1;i<=9;i++)list.add(i);//将这三个数的所有组成从list中去除，list为空，则返回tureif (remove(a,list)&&remove(b,list)&&remove(c,list)&&list.size()==0)flag = true;return flag;}//看list中是否包含该数的组成数，如果有，从list中去除，返回true。static boolean remove(int number,List<Integer> list){char[] numberChar = String.valueOf(number).toCharArray();for(char ch:numberChar){int tempNumber = ch-'0';if(list.size()>0&&list.contains(tempNumber)){list.remove(Integer.valueOf(tempNumber));}else{return false;}}return true;}public static void main(String[] args) {long beginTime = System.currentTimeMillis();int sum = 0;//相除for(int i=1000;i<=9999;i++){for(int j=2;j<=4999;j++){if(i%j==0){int k = i/j;if(isPandigital(i,j,k)){System.out.println("i:"+i+" j:"+j+" k:"+k);sum += i;break;}}}}/** 相乘 //640msfor(int i=2;i<=99;i++){for(int j=100;j<=4999;j++){int k = i*j;if(isPandigital(k,i,j)){System.out.println("i:"+i+" j:"+j+" k:"+k);sum += k;break;}}}*/System.out.println("sum==="+sum);//45228long endTime = System.currentTimeMillis();System.out.println(endTime-beginTime+"ms");//800ms}}