UVA-10591 Happy Number

Happy number

UVA-10591

Let the sum of the square of the digits of a positive integer S0 is represented by S1. In a similar-a-to, let the sum of the squares of the digits of S1 is represented by S2 and so on. If Si = 1 for some i≥1 and then the original integer S0 are said to be Happy number. A number, which is isn't happy, is called unhappy number. For example 7 is a Happy number since 7→49→97→130→10→1 and 4 are an unhappy number since 4→16→37→58→89→ 145→42→20→4.

Input

The input consists of several test cases, the number of which you is given in the first line of the input. Each test case consists the one line containing a single positive integer N smaller than 109.

Output

For each test case, you must print one of the following messages:

Case #p: N is a Happy number.

Case #p: N was an unhappy number.

Here P stands is the case number (starting from 1).

You should print the first message if the number N is a happy number. Otherwise, print the second line.

Sample Input

3

7

4

13

Sample Output

Case #1:7 is a Happy number.

Case #2:4 are an unhappy number.

Case #3: Happy number.

The problem of recursion or cycle I feel very good, to tell the truth first time I did not understand the problem, is to give you a number, if the number of each number of square sum is 1 is happy, manually tried, less than 10 of only 1 satisfied, so I just put it squared and to a number on the line, Otherwise the loop will not go out, and then we can continue the loop recursion.

#include <stdio.h>intFintb) {    ints=0;  while(b) {intt=b%Ten; S+=t*T; b/=Ten; }    if(s<Ten)returns; Else returnf (s);}intMain () {intT; scanf ("%d",&T);  for(intI=1; i<=t; i++)    {        intx; scanf ("%d",&x); printf ("Case #%d:%d is a", i,x); if(f (x) = =1) printf ("Happy number.\n"); Elseprintf"N Unhappy number.\n"); }    return 0;}

UVA-10591 Happy Number