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 <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\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;
}

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.

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