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 S1. In a similar way, let the sum of the squares of the digits of S1 be represented by S2 and so on. If Si = 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：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;
}
}
}