問題描述:
Take the number 192 and multiply it by each of 1, 2, and 3:
192 * 1 = 192
192 * 2 = 384
192 * 3 = 576
By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n 1?
解決問題:
public class Problem38 {
public static boolean IsNumber(int number){
StringBuffer result = new StringBuffer();
int mul = 1;
while(result.length()<9){
String tmp = number*mul+"";
result.append(tmp);
mul ++;
}
if(result.length()!=9){
return false;
}else{
boolean[] elements = new boolean[10];
Arrays.fill(elements, true);
int r = Integer.parseInt(result.toString());
System.out.println("Result:"+r);
while(r!=0){
int cur = r%10;
if(cur==0||!elements[cur]){
return false;
}
elements[cur] = false;
r = r/10;
}
return true;
}
}
public static void main(String[] args){
for(int i=9999; i>0; i--){
if(IsNumber(i)){
System.out.println(i);
break;
}
}
System.out.println(IsNumber(9327));
}
}