UVA - 10591 Happy Number（hash）

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� 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.

 

题目大意：所谓的Happy数字，就是给一个正数s，然后计算它每一个位上的平方和，得到它的下一个数， 然后下一个数继续及选每位上的平方和……如果一直算下去，没有出现过之前有出现过的数字而出现了1，那么恭喜，这就是个Happy Number，如果算下去的过程中出现了一个之前出现过的，那么就不Happy了。

解析：直接用set进行判重

#include <cstdio>#include <cstring>#include <set>using namespace std;int main() {int t;int cas = 1;int num,sum;set<int> vis;scanf("%d",&t);while(t--) {vis.clear();scanf("%d",&num);int now = num;while(1) {if(now == 1) {printf("Case #%d: %d is a Happy number.\n",cas++,num);break;}if(vis.count(now)) {printf("Case #%d: %d is an Unhappy number.\n",cas++,num);break;}vis.insert(now);sum = 0;while(now) {int tmp = now % 10;sum += tmp * tmp;now /= 10;}now = sum;}}return 0;}