如何有效地在Java中生成Perfect数位不变数字？

Question

A PDDI is a number such the the sum of all the digits raised to themselves is equal to the number itself.

For example, 3435 = (3^3) + (4^4) + (3^3) + (5^5)

The code below takes too long to check for PDDIs between one to a huge number. Is there any way to make it faster?

    System.out.print("Enter the number");
    Scanner s = new Scanner(System.in);
    int n = s.nextInt();

    int m = 0, sum = 0, k = 0;
    // We're going to try all integers between one to n.
    for(int i = 1; i<=n; i++){
       sum = 0;
       m = i;
       while(m>0){
          k = m % 10;
          sum = sum + (int)Math.pow(k, k);
          m = m/10;
       }

       if(i == sum)
          System.out.println(i);
    }

Answer 1

The number from 0 to 9 to the power of 2 can be precalculated and kept in a an array.

int powered [] = new int [10];
powered[0] = 0;
powered[1] = 1;
powered[2] = 4;
..
powered[9] = 81;

Then for each digit fech the powered number using the digit as an index to the powered array.

For example 234 would be powered[2] + powered[3] + powered[4]

This will save some math operations.

Also you could think of a multithreaded approach having N threads doing the calculations for different numbers in parallel.

Answer 2

Use cached values instead of Math.pow

Since you are using only power from 0 to 9, you could cache these values in a int[] instead of computing Math.pow(k, k) everytime. It won't improve that much but, it's a start.

int[] pows = new int[] {0, 1, 4, 27, 256, 3125, 46656, 823543, 16777216, 387420489 };
for (int i = 0; i < 10; ++i) {
    pows[i] = (int) Math.pow(i, i);
}

System.out.print("Enter the number");
Scanner s = new Scanner(System.in);
int n = s.nextInt();

int m = 0, sum = 0, k = 0;
// We're going to try all integers between one to n.
for(int i = 1, i<=n, i++){
   sum = 0;
   m = i;
   while(m>0){
      k = m % 10;
      sum = sum + pows[k]; // use cached values here
      m = m/10;
   }

   if(i == sum)
      System.out.println(i);
}

Skip useless values

Based on logic you may skip some iterations. Lets take the number 281 as an example, which gives 4 + 16777216 + 1 = 16777251 , the result is above 281, so theres no changes that 282, 283, 284, ... 289 gives a number equals to 281. In such cases you may want to skip these useless iterations by manually incrementing i .

int[] pows = new int [10];
for (int i = 0; i < 10; ++i) {
    pows[i] = (int) Math.pow(i, i);
}

System.out.print("Enter the number");
Scanner s = new Scanner(System.in);
int n = s.nextInt();

int m = 0, sum = 0, k = 0, lastNumberDigit;
// We're going to try all integers between one to n.
for(int i = 1, i<=n, i++){
   sum = 0;
   m = i;
   while(m>0){
      lastNumberDigit = m; // on the last iteration, we'll get the last digit
      k = m % 10;
      sum = sum + pows[k]; // use cached values here
      m = m/10;
   }

   if(i == sum) {
      System.out.println(i);
   } else if (sum > i) {
      i += (10 - lastNumberDigit - 1); // jump to the next decade (-1 because the for will apply i++ on it)
   }
}

I used this logic on decade, but you may want to extend it to hundreds or even more, but it will be much more tricky.