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.

* 根據題意分析：
* 因爲A*B=C，而且A、B不能等於1，所以C>B,C>A。
* A的位數+B的位數+C的位數=9，所以C大於3位，C>=1000。
* A的位數+B的位數=9-C的位數，若C爲4位，則有兩種組合(A爲1位，B爲4位；A爲2位，B爲3位)可以滿足。
* 若C爲5位，則A的位數+B的位數=4，不可能滿足A*B=C的條件。
*
* 所以C的範圍爲（1000-9999）
* A/B的範圍爲(2-4999),因爲A、B位置可互換，
* 所以A的取值範圍可確定(2-99),B的取值範圍(100-4999)

 

 

import java.util.ArrayList; import java.util.List; public class Problem32 { //判斷這三個數能否組成1-9的全數碼。利用list來比較，list初始化了1-9的數字。 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); //將這三個數的所有組成從list中去除，list爲空，則返回ture if (remove(a,list)&&remove(b,list)&&remove(c,list)&&list.size()==0) flag = true; return flag; } //看list中是否包含該數的組成數，如果有，從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 //640ms for(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);//45228 long endTime = System.currentTimeMillis(); System.out.println(endTime-beginTime+"ms");//800ms } }