UVA 10591 Happy Number

Problem C

Happy Number

Time Limit

1 Second

 

Let the sum of the square of the digits of a positive integer S₀ be represented by S_1. In a similar way, let the sum of the squares of the digits of S₁ be represented by S₂ and so on. If S_i = 1 for some i ³ 1, then the original integer S₀ is said to be Happy number. A number, which is not happy, is called Unhappy number. For example 7 is a Happy number since 7 -> 49 -> 97 -> 130 -> 10 -> 1 and 4 is 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 are given in the first line of the input. Each test case consists of 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 is an Unhappy number.

 

Here p stands for 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

Output for Sample Input

3

7

4

13

Case #1: 7 is a Happy number.

Case #2: 4 is an Unhappy number.

Case #3: 13 is a Happy number.

题意：输入一个数字n。每次都把这个数字进行变换：把该数字变换成该数字每个位上数字的平方和。。如果变换过程中。出现重复的。这个数字就是Unhappy的。如果变换过程中出现1，这个数字就是Happy的。。

思路：完全可以用一个vis数组来标记遇到过的数字。。虽然题目中N最大可以为10^9。。但是经过第一次变换之后。。999999999的平方和是最大的。。为9 ^ 2 * 9 大概是700多 。。

所以vis只要开 1000就可以了。。。 然后进行模拟。。每次变换完进行判断。。注意：如果用这个方法。判断的时候 还要跟原来的n进行一次比较。

#include <stdio.h>#include <string.h>int t;int n;int vis[1000];int judge;int main(){    scanf("%d", &t);    int tt = 1;    while (t --)    {memset(vis, 0 ,sizeof(vis));scanf("%d", &n);int sb = n;while (1){    int sum = 0;    while (sb)    {sum += (sb % 10) * (sb % 10);sb /= 10;    }    sb = sum;    if (sum == 1)    {judge = 0;break;    }    if (vis[sum] == 1 || sum == n)    {judge = 1;break;    }    else    {vis[sum] = 1;    }}if (judge == 0)    printf("Case #%d: %d is a Happy number.\n", tt ++, n);if (judge == 1)    printf("Case #%d: %d is an Unhappy number.\n", tt ++, n);    }    return 0;}