[Blue Bridge Cup] [improve the algorithm] stupid monkey

Problem Description

Description
Vocabulary stupid monkey is very small, so every time when English multiple-choice questions are a headache.
But he found a way, the test proved that in this way the chance to choose the option when selected for very large!
This method is specifically described as follows: Suppose MAXN largest number of occurrences is the number of letters in a word appear, Minn number of occurrences is the least number of occurrences of the letter word, if maxn-minn is a prime number, it is considered stupid monkey a Lucky Word, this word is probably the right answer.
Input
Input file only one line, is a word which may only appear in lowercase letters, and the length is less than 100.
Output
A total of two lines of output file:
The first line is a string, assuming that the input word is Lucky Word, then the output "Lucky Word", otherwise output "No Answer";
the second line is an integer, if the input word is Lucky Word the output maxn-minn value, otherwise output 0.
Sample Input
error
Sample Output
Lucky Word
2

Thinking problems

• 1 using a first length of the array 26, the number of each letter appears recorded
• 2 to find the maximum and minimum values ​​from the array
• Analyzing maxn-minn 3 is prime

!! Precautions

• 0 and 1 is not a prime number
• "A" and "aabb" These two cases should special note

The complete code

#include <iostream>
#include <string>
using namespace std;

bool issuper(int num){
    // !!!0和1都不是质数
    if(num == 0){
        return false;
    }else if(num == 1){
        return false;
    }else{
        for(int i=2;i<num/2;i++){
            if(num%i==0){
                return false;
            }
        }
    }
    return true;
}

int main()
{
    string c;
    int len;
    int ll[26] = {0};
    int maxn=-1,minn=105;
    int judge;

    c = "aaabbb";
    cin >> c;
    len = c.length();

    // 创建一个次数表
    for(int i=0;i<len;i++){
        ll[c[i]-'a']++;
    }
    // 如果是只有一类字母的情况
    // 遍历次数表找出最大最小值
    for(int i=0;i<26;i++){
        if(ll[i]==0) continue;
        if(ll[i]>maxn){
            maxn = ll[i];
        }
        if(ll[i]<minn){
            minn = ll[i];
        }
    }
    if(len == maxn) minn=0;
    // 判断是不是质数
    bool ans = issuper(maxn-minn);
    if(!ans){
        cout << "No Answer" << endl;
        cout << 0 << endl;
    }else{
        cout << "Lucky Word" << endl;
        cout << maxn-minn << endl;
    }

    return 0;
}