All in one 1407: stupid monkey

【 Title Description 】

Stupid little monkey has a small vocabulary , So every time I do multiple choice questions, I have a headache . But he found a way , It has been proved that , It's a very good way to choose the right option ！

The specific description of this method is as follows ： hypothesis maxnmaxn It's the number of letters that appear most frequently in a word ,minnminn It's the number of letters that appear the least in a word , If maxn−minnmaxn−minn It's a prime number , So stupid little monkey thinks it's a Lucky WordLucky Word, Such a word is probably the right answer .

【 Input 】

There is only one line , It's a word , Only lowercase letters are possible , And the length is less than 100100.

【 Output 】

There are two lines , The first line is a string , Suppose the input word is Lucky WordLucky Word, Then output “Lucky WordLucky Word”, Otherwise output “No AnswerNo Answer”;

The second line is an integer , If the input word is Lucky WordLucky Word, Output maxn−minnmaxn−minn Value , Otherwise output 00.

【 sample input 】

error

【 sample output 】

Lucky Word
2

【 Tips 】

The sample input :

Examples #2：

olympic

Sample output :

Examples #2：

No Answer 0

#include<stdio.h>
#include<string.h>
int main()
{
	char s[120];
	scanf("%s", s);
	int len = strlen(s);
	int i;
	int a;
	int b[150] = { 0 };
	for (i = 0; i < len; i++)
	{
		a = s[i];
		b[a]++;
	}
	int max=0;
	int min=101;
	for (i = 0; i < 150; i++)
	{
		if (b[i] != 0)
		{
			if (max < b[i])
				max = b[i];
			if (min > b[i])
				min = b[i];
		}
	}
	int num = max - min;
	if (num < 2)
	{
		printf("No Answer\n");
		printf("0");
	}
	if (num >= 2)
	{
		for (i = 2; i <= num; i++)
		{
			if (num % i == 0)
			{
				break;
			}
		}
		if (num == i)
		{
			printf("Lucky Word\n");
			printf("%d", num);
		}
		else
		{
			printf("No Answer\n");
			printf("0");
		}
	}
	return 0;
}