Problem 32

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);
	}