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[英]is it possible to encrypt 10 digit number to 4 digit number using any algorithm?
[英]Algorithm involving figuring out 10 digit number puzzle
要確定數字是否可以被7整除,請從數字中減去最后一位數字,將其加倍,然后從剩余數字中減去加倍的數字。 如果結果可被7整除(例如14、7、0,-7等),則數字可被7整除。 這可能需要重復幾次。 示例:3101被7整除嗎?
310 - take off the last digit of the number which was 1
-2 - double the removed digit and subtract it
308 - repeat the process by taking off the 8
-16 - and doubling it to get 16 which is subtracted
14 - the result is 14 which is a multiple of 7
以下是我用來獲取號碼的代碼:
for(int O =0; O <= 9 ; O++) {
String a = String.valueOf(number[0]);
String b = String.valueOf(number[1]);
String c = String.valueOf(number[2]);
String d = String.valueOf(number[3]);
String e = String.valueOf(number[4]);
String f = String.valueOf(number[5]);
String h = a+b+c+d+e+f;
int abcdef = Integer.valueOf(h);
if ( (abcdef -(2*O) % 7) ==0 )
number [6] = O;
}
但是,它並沒有給我這樣的數字,我能夠得到一個直到6的數字,直到每個數字都可以被相應的索引整除(如果我以1,而不是0開頭)。這意味着索引1可以被1整除,索引2可以被2整除,索引3可以被3整除,......索引7可以被7整除。我想形成這樣的數字。請注意,我可以通過以下方式無需使用算法即可完成此操作:
for(int O =0; O <= 9 ; O++) {
String a = String.valueOf(number[0]);
String b = String.valueOf(number[1]);
String c = String.valueOf(number[2]);
String d = String.valueOf(number[3]);
String e = String.valueOf(number[4]);
String f = String.valueOf(number[5]);
String g = String.valueOf(O);
String h = a+b+c+d+e+f+g;
int abcdefg = Integer.valueOf(h);
if ( (abcdefg % 7) ==0 )
number [6] = O;
}
但是,我真的很想使用我在開始時描述的算法。
調用代碼示例 :
int [] num = new int [7];
for (int i = 1000000; i < 9999999; i++)
{
// put i into the array and check it
if (checkDigitsDivisible(i, num))
{
System.out.println(i);
}
}
檢查一個數字的前幾位數是否可以被1整除,前兩位數可以被2除數,前3位可以被3除數,依此類推。
public static boolean checkDigitsDivisible (int num, int [] arr)
{
int one = num / 1000000;
int two = num / 100000;
int three = num / 10000;
int four = num / 1000;
int five = num / 100;
int six = num / 10;
int seven = num;
arr[0] = one % 10;
arr[1] = two % 10;
arr[2] = three % 10;
arr[3] = four % 10;
arr[4] = five % 10;
arr[5] = six % 10;
arr[6] = seven % 10;
return (one % 1 == 0) &&
(two % 2 == 0) &&
(three % 3 == 0) &&
(four % 4 == 0) &&
(five % 5 == 0) &&
(six % 6 == 0) &&
(isDivisibleBy7(seven));
}
檢查數字是否可被7整除的遞歸解決方案(問題中所述的算法):
public static boolean isDivisibleBy7 (int n)
{
n = Math.abs(n);
if (n == 0 || n == 7 || n == 14)
{
return true;
}
else
{
int lastDigit = n % 10;
int lastDigitOff = n / 10;
int remainder = lastDigitOff - (lastDigit * 2);
remainder = Math.abs(remainder);
if (remainder > n)
{
return false;
}
return isDivisibleBy7(remainder);
}
}
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