问题描述：
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.
解决问题：
public static void main(String[] args) {
System.out.println(sum());
int result = 0;
for (int i = 1000; i < 100000; i++) {
boolean[] matcher = new boolean[10];
Arrays.fill(matcher, false);
int middle = (int) Math.sqrt(i);
boolean ok = true;
for(int j=2; j<middle&&ok;j++){
Arrays.fill(matcher, false);
if(i%j==0){
int k = i/j;
int tmp = i;
while(tmp!=0){
if(matcher[tmp%10]==true||tmp%10==0){
ok = false;
break;
}
matcher[tmp%10] = true;
tmp = tmp/10;
}
if(ok){
tmp = j;
while(tmp!=0){
if(matcher[tmp%10]==true||tmp%10==0){
ok = false;
break;
}
matcher[tmp%10] = true;
tmp = tmp/10;
}
}
if(ok){
tmp = k;
while(tmp!=0){
if(matcher[tmp%10]==true||tmp%10==0){
ok = false;
break;
}
matcher[tmp%10] = true;
tmp = tmp/10;
}
}
if(ok){
boolean find = true;
for(int l=1; l<10; l++){
if(matcher[l]==false){
find = false;
break;
}
}
if(find ){
System.out.println("i:"+i+",j:"+j+",k:"+k);
result += i;
break;
}
}
}
ok = true;
}
}
System.out.println(result);
}