1406357289是一個pandigital數,因爲它包含了0到9之間每個數字且只包含了一次。此外它還有一個有趣的子串整除性質。
令d1表示其第一位數字,d2表示第二位,以此類推。這樣我們可以得到:- d2d3d4=406 能被 2 整除
- d3d4d5=063 能被 3 整除
- d4d5d6=635 能被 5 整除
- d5d6d7=357 能被 7 整除
- d6d7d8=572 能被 11 整除
- d7d8d9=728 能被 13 整除
- d8d9d10=289 能被 17 整除
求所有具有如上性質的0到9pandigital數。
import java.util.Arrays;
public class Problem43
{
public static void main(String[] args)
{
long start = System.currentTimeMillis();
System.out.print("answer: ");
howmany();
long end = System.currentTimeMillis();
System.out.print("time: ");
System.out.println(end - start);
}
static int array[] = new int[10];
static void howmany()
{
int qiuyu[] = {2,3,5,7,11,13,17};
for (int i = 100; i <= 999; i++)
{
array[0] = i / 100;
array[1] = i / 10 % 10;
array[2] = i % 10;
zuhe(i, qiuyu, 0);
}
System.out.println(answer);
}
static long answer = 0;
static void zuhe(int n, int qiuyu[], int k)
{
if (k == 7)
{
if (ispandi())
{
long s = 0;
for (int i = 0; i < 10; i++)
{
s = s * 10 + array[i];
}
answer += s;
}
return;
}
for (int i = 0; i <= 9; i++)
{
int temp = (n % 100 * 10 + i);
if ( temp % qiuyu[k] == 0 )
{
array[k + 3] = temp % 10;
zuhe(temp, qiuyu, k + 1);
}
}
}
static boolean ispandi()
{
int ch[] = Arrays.copyOf(array, 10);
Arrays.sort(ch);
for (int i = 0; i < ch.length; i++)
{
if (ch[i] != i)
{
return false;
}
}
return true;
}
}
answer: 16695334890
time: 143