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

ANSWER(with a little presentation error)

#include 
    
     #include 
     
      //I think I should improve my POOR English, so all the comments are written in Englishint 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\n", 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;}