三個三位數,第一加第二等於第三,三個數所有各個位的數都不相同(1-9),第二個數是第一個數的兩倍,求三個數

 

public class Test {

    public static void main(String[] args) {
        int arr[] = new int[1000];
        int a, b, c, d, e, f, k, i;
        int x, y, z, o, p, q, r, s, t;
        for (i = 123, k = 0; i <= 987; i++) {
            a = i % 10;
            b = i / 10 % 10;
            c = i / 100;
            if (a != b && b != c && a != c && b != 0 && a != 0) {  //求得一個三位數的各個位數都不相同,並存到數組中
                arr[k] = i;
                k++;
            }
        }
        int sum = 1;
        for (int m = 1; m <= 9; m++)  //求得9!
            sum = sum * m;
//System.out.println(sum);

        for (d = 0; d <= arr.length - 1 && arr[d] != 0; d++) {
            o = arr[d] % 10;
            p = arr[d] / 10 % 10;
            q = arr[d] / 100;
            for (e = 0; e <= arr.length - 1 && arr[e] != 0; e++) {
                r = arr[e] % 10;
                s = arr[e] / 10 % 10;
                t = arr[e] / 100;
                if(arr[d]==arr[e]) continue;  //爲減少循環的次數,當第一個數與第二個數相同時,繼續下一層循環
                for (f = 0; f <= arr.length - 1 && arr[f] != 0; f++) {
                    x = arr[f] % 10;
                    y = arr[f] / 10 % 10;
                    z = arr[f] / 100;
                    if(arr[f]==arr[e]||arr[f]==arr[d]) continue;//同上
                    if (((arr[d] + arr[e]) == arr[f]) && (arr[e] == 2 * arr[d])
                            && (o * p * q * s * r * t * x * y * z == sum))
                        System.out.println(arr[d] + " " + arr[e] + " " + arr[f]);
                }
            }
        }

    }

}