UVa944 - Happy Numbers(动态规划)

Let the sum of the squares of the digits of a positive integer s0 be represented by s1. In a similar way,
let the sum of the squares of the digits of s1 be represented by s2, and so on. If si = 1 for some i  1,
then the original integer s0 is said to be happy. For example, starting with 7 gives the sequence
7; 49(= 7 ^ 2); 97(= 4 ^ 2 + 9 ^ 2); 130(= 9 ^ 2 + 7 ^ 2); 10(= 1 ^ 2 + 3 ^ 2); 1(= 1 ^ 2);
so 7 is a happy number.
The rst few happy numbers are 1, 7, 10, 13, 19, 23, 28, 31, 32, 44, 49, 68, 70, 79, 82, 86, 91, 94,
97, 100, : : : The number of iterations i required for these to reach 1 are, respectively, 1, 6, 2, 3, 5, 4, 4,
3, 4, 5, 5, 3, : : :
A number that is not happy is called unhappy. Once it is known whether a number is happy
(unhappy), then any number in the sequence s1; s2; s3; : : : will also be happy (unhappy). Unhappy
numbers have eventually periodic sequences of si which do not reach 1 (e.g., 4, 16, 37, 58, 89, 145, 42,
20, 4, : : :).
Any permutation of the digits of a happy (unhappy) number must also be happy (unhappy). This
follows from the fact that addition is commutative. Moreover, the product of a happy (unhappy)
number by any power of ten is a happy (unhappy) number. Example: 58 is an unhappy number; then,
so are 85, 580, 850, 508, 805, 5800, 5080, 5008, 8050, 8500, and so on.
Decide which numbers, in a given closed interval, are happy numbers.
Input
The input has n lines each of them corresponding to test case. Every line contains two positive integers
between 1 and 99999 each; the rst integer, L, is the low limit of the closed interval; the second one,
H, is the high limit (L  H).
Output
The output is composed of the happy numbers that lie in the interval [L; H], together with the number
of iterations required for the corresponding sequences of squares to reach 1.
There must be a line for each happy number containing the happy number followed by a space and
the number of iterations required for the sequence of squares to reach 1.
Print a blank line between two consecutive test cases.
Note: The de nition of happy numbers is from MathWorld - http://mathworld.wolfram.com/
Sample Input
5 28
233 250
Sample Output
7 6
10 2
13 3
19 5
23 4
28 4
236 6
239 6

import java.io.FileInputStream;import java.io.BufferedInputStream;import java.io.OutputStreamWriter;import java.io.PrintWriter;import java.util.Scanner;import java.util.Arrays;import java.util.HashSet;public class Main implements Runnable{private static final boolean DEBUG = false;private static final int N = 100000;private PrintWriter cout;private Scanner cin;private int l, h;private int cas = 1;private int[] f;private HashSet<Integer> hs = new HashSet<Integer>();private void init(){try {if (DEBUG) {cin = new Scanner(new BufferedInputStream(new FileInputStream("d:\\OJ\\uva_in.txt")));} else {cin = new Scanner(new BufferedInputStream(System.in));}cout = new PrintWriter(new OutputStreamWriter(System.out));f = new int[N];Arrays.fill(f, -1);} catch (Exception e) {e.printStackTrace();}}private boolean input(){if (!cin.hasNextInt()) return false;l = cin.nextInt();h = cin.nextInt();return true;}private int dp(int x){if (f[x] != -1) return f[x];if (x == 1) return 1;int sum = 0;int tmp = x;while (tmp != 0) {int rem = tmp % 10;sum += rem * rem;tmp /= 10;}if (hs.contains(sum)) return f[x] = 0;else {hs.add(sum);tmp = dp(sum);if (tmp == 0) return f[x] = 0;else return f[x] = 1 + tmp;}}private void solve(){if (cas++ > 1) {cout.println();}for (int i = l; i <= h; i++) {hs.clear();int tmp = dp(i);if (tmp > 0) {cout.println(i + " " + tmp);}}cout.flush();}public void run(){init();while (input()) {solve();}}public static void main(String[] args){new Thread(new Main()).start();}}