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.
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.
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 8Sample Output
Lucky number! Lucky number!
-----------------------------------------------------------------
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;
}