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.

InputThe first line contains an integer stands for the number of test cases.
Each test case contains an integer n (n >= 0).
OutputFor each case, output “Lucky number!” if the number is lucky, otherwise output “What a pity!”.Sample Input

2
0
8

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;
}