An Easy C Program Problem

找幸运数

题目描述

数字8最多的那个数为幸运数。
输入n和n个整数，找这n个数中的幸运数。在主函数中调用ndigit函数，判断某个整数x含数字8的个数。如果有多个幸运数输出第一个幸运数，如果所有的数中都没有含数字8，则输出NO.

函数int ndigit（int n,int k)功能：统计整数n中含数字k的个数。

输入描述

输入n个n个整数

输出描述

幸运数

输入样例

5 568 567 328 48768 8688

输出样例

8688

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ANSWER(with a little presentation error)

#include <stdio.h>
#include <stdlib.h>

//I think I should improve my POOR English, so all the comments are written in English

int ndigit (int n, int k);

int main()
{
	/**
	 * @param n INPUT 1
	 * @param num the temp of the number in INPUT
	 * @param luckyNum the lucky number
	 * @param luckyDigCount the count of lucky digit in the lucky number
	 */
	int n, i, num, luckyNum = 0, luckyDigCount = 0;

	//get the INPUT
	scanf("%d", &n);

	//get n numbers from console 
	//and find the lucky number
	for (i = 0; i < n; i++)
	{
		//get the input
		scanf("%d", &num);

		//if the count of lucky digit in current number more than current lucky number's
		if (ndigit(num, 8) > luckyDigCount)
		{
			//set current number as lucky number
			luckyDigCount=ndigit(num,8);	
			luckyNum = num;
		}
	}
	//if lucky number doesn't have a lucky digit
	//that means there is no lucky number in this test case
	//so, Print "NO"
	if (luckyDigCount==0)
	{
		printf("NO");
	}
	else
	{
		//Print the lucky number 
		printf("%d
", luckyNum);
	}
}
/**
 * get the count of lucky digit in the param n
 * @param  n test number
 * @param  k lucky digit
 * @return   the count of lucky digit in the param n 
 */
int ndigit (int n, int k)
{
	int count = 0;
	for (; n; n /= 10)
	{
		if (n%10 == k)
		{
			count++;
		}
	}

	return count;
}

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SUMMARY

What if the OUTPUT is the biggest lucky number?
Add a judgement statement,that compare current number to the previous lucky number, after we ensure current number is one of the lucky numbers.