uva 10591 Happy Number（判重)

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description：

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 S₁_. 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

Case #1: 7 is a Happy number.

Case #2: 4 is an Unhappy number.

Case #3: 13 is a Happy number.

Problemsetter: Mohammed Shamsul Alam

题目大意：根据所给的一个数，每次都取它的每个位上的数的平方和作为下一个数，如果可以推出1的话就是happy number，如果出现相同的就是 unhappy number。

解题思路：因为数字不超过10^9，所以这个数所形成的下一个数最大是 9 * 81 = 729；

而前面如果给出的数大于这个数，说明以后也不会出现这个数，所以可以不用记录进去。开个729的数组就可以完成判重了。

#include<stdio.h>#include<string.h>const int N = 800;int visit[N];int n;int main(){scanf("%d", &n);for(int i = 1; i <= n; i++){int num;memset(visit, 0, sizeof(visit));scanf("%d", &num);if(num == 1){printf("Case #%d: %d is a Happy number.\n", i, num);continue;}if(num <= 729)visit[num] = 1;int x = num;while(1) {int dight = 0; while(num){ dight += (num % 10) * (num % 10);num = num / 10;}if(dight == 1){printf("Case #%d: %d is a Happy number.\n", i, x);break;}if(visit[dight]){printf("Case #%d: %d is an Unhappy number.\n", i, x);break;}else visit[dight] = 1;num = dight;}}}

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