Lucky Number

2019-11-06 06:28:37

字體：大中小

供稿：網(wǎng)友

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

208Sample OutputLucky number!Lucky number!-----------------------------------------------------------------思路分析：輸入一個(gè)數(shù)，對這個(gè)數(shù)進(jìn)行三種情況的判斷，1、該數(shù)是否是8的倍數(shù)；2、該數(shù)每一位的和是否是8的倍數(shù)；3、該數(shù)每一位的平方的和是否是8的倍數(shù)；如果滿足，則輸出Lucky number！否則輸出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的倍數(shù)，和是八得倍數(shù)，和的平方是八的倍數(shù) int digsum(int a){//和是否是8的倍數(shù)	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的倍數(shù){	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;}