Lucky Number

   To Chinese people, 8 is a lucky number. Now your task is to judge if a number is lucky.
We say a number is lucky if it’s a multiple of 8, or the sum of digits that make up the number is a multiple of 8, or the sum of every digit’s square is a multiple of 8.

Input

The first line contains an integer stands for the number of test cases.
Each test case contains an integer n (n >= 0).

Output

For each case, output “Lucky number!” if the number is lucky, otherwise output “What a pity!”.

Sample Input

208

Sample Output

Lucky number!Lucky number!

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思路分析：

输入一个数，对这个数进行三种情况的判断，

1、该数是否是8的倍数；

2、该数每一位的和是否是8的倍数；

3、该数每一位的平方的和是否是8的倍数；

如果满足，则输出Lucky number！否则输出What a pity!

#include <iostream>#include<string>#include<stdio.h>#include<cmath>/* run this program using the console pauser or add your own getch, system("pause") or input loop */using namespace std;//8的倍数，和是八得倍数，和的平方是八的倍数 int digsum(int a){//和是否是8的倍数int sum=0;while(a!=0){sum+=a%10;a=a/10;}return sum;}int sumf(int a)  //和的平方是否是8的倍数{int sum=0;while(a!=0){sum=sum+pow(a%10,2);a=a/10;}return sum;}int main(int argc, char** argv) {int n,num[1000],sum;cin>>n;for(int i=0;i<n;i++){cin>>num[i];if(num[i]==0)printf("Lucky number!\n");else if(num[i]%8==0){printf("Lucky number!\n");}else if((sum=digsum(num[i]))%8==0){printf("Lucky number!\n");}else if((sum=sumf(num[i]))%8==0){printf("Lucky number!\n");}else{printf("What a pity!\n");} }  return 0;}