uva 10591 Happy Number(判重）

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 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.

 

Problemsetter: Mohammed Shamsul Alam

题目大意:给出一个数字，对应一个变换，如果最终变换的结果可以变成1,那么这个数解释happy number，否则就是unhappy number，对应变换为数的每个位数上的数值的平方的和为新的数值。

解题思路：对应任何一个数，经过变换后的值最大为9 * 9 * 9， 所以开一个1000的数组就可以将所有情况标记（注意开始输入的值没有必要标记，也开不了那么大的数组），用一个循环对数进行变换，碰到已经出现过的数值就可以跳出，跳出后判断1是否遍历即可（注意输出时a和an).

#include <stdio.h>#include <string.h>const int N = 10005;int count(int cur) {    int sum = 0, t;    while (cur){    t = cur % 10;sum = sum + t * t;cur = cur / 10;    }    return sum;}int main() {    int cas = 1, n, m, cur;    int v[N];    scanf("%d", &m);    while (m--) {scanf("%d", &n);memset(v, 0, sizeof(v));cur = n;while (1) {    cur = count(cur);    if (v[cur])break;    v[cur] = 1;}printf("Case #%d: %d is %s number.\n", cas++, n, v[1]?"a Happy":"an Unhappy");    }    return 0;}