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
International Islamic University Chittagong
Special thanks to Muhammad Abul Hasan
99999999->81*8
标记数组判循环最大9*9*8
#include<stdio.h> #include<string.h> int sort(int x) {int y=0; while (x>0) {y=y+(x%10)*(x%10); x=x/10; } return y; } int main() {int n,m,t,i,visit[1000]; scanf("%d",&t); for (i=1;i<=t;i++) {scanf("%d",&n); memset(visit,0,sizeof(visit)); m=n; while (n) {m=sort(m); if (m==1) {printf("Case #%d: %d is a Happy number.\n",i,n);break;} if (visit[m]) {printf("Case #%d: %d is an Unhappy number.\n",i,n);break;} visit[m]=1; } } return 0; }