android 手机屏幕解锁最多有多少种？

#python

from itertools import *

impossible = {‘13‘: ‘2‘, ‘46‘: ‘5‘,  ‘79‘: ‘8‘, ‘17‘: ‘4‘, ‘28‘: ‘5‘, ‘39‘: ‘6‘,  ‘19‘: ‘5‘,  ‘37‘: ‘5‘,‘31‘: ‘2‘, ‘64‘: ‘5‘,‘97‘: ‘8‘,‘71‘: ‘4‘,‘82‘: ‘5‘, ‘93‘: ‘6‘,‘91‘: ‘5‘,‘73‘: ‘5‘}

def count_unlock_screen():
    bucket = chain(*(permutations(‘123456789‘, i) for i in range(4, 10)))
    count = 0
    for i in bucket:
        striter = ‘‘.join(i)
        for key, value in impossible.items():
            if key in striter and value not in striter[:striter.find(key)]:
                break
        else:
            count += 1
    return count

print( count_unlock_screen() )

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

bool legal(string left, string right, string mid, string &a)
{    
    string test1 = left + right;
    string test2 = right + left;
    string::size_type p1 = a.find(test1);
    string::size_type p2 = a.find(test2);
    string::size_type q = a.find(mid);
    
    if ((p1 != string::npos) && (q == string::npos))
        return false;
    if ((p2 != string::npos) && (q == string::npos))
        return false;
    if((p1 != string::npos) && q > p1)
        return false;
    if ((p2 != string::npos) && q > p2)
        return false;
    return true;
}


bool is_legal(string &a)
{
    string::size_type p = string::npos;
    string one("1");string two("2");string three("3");
    string four("4"); string five("5"); string six("6");
    string seven("7"); string eight("8"); string nine("9");
    if ((!legal(one, three, two, a)) || (!legal(one, seven, four, a)) || (!legal(one, nine, five, a)) || (!legal(three, seven, five, a)) || (!legal(three, nine, six, a)) || (!legal(seven, nine, eight, a))||(!legal(two, eight, five, a))||(!legal(four, six, five, a)))
        return false;
    else return true;
}

void swap(char &index1, char &index2)
{
    char temp = index1;
    index1 = index2;
    index2 = temp;
}


//全排列中的下一个
bool next_permutation(string &a)
{
    int i = 0;
    int size = a.size();
    for (i = size - 2; i >= 0; i--)            //从size开始找到第一个满足a[i] < a[i+1]的位置i
    {
        if (a[i] < a[i + 1])
            break;
    }
    if (i == -1)                            //寻找结束后若i=-1说明当前的a数组已经达到字典序的最大
    {
        return true;
    }
    int j = i + 1;                            //j用于遍历i---n之间的元素
    int min_larger = j;                       //min_larger存储i---n之间满足  a[min_larger] > a[j] 中的最小者，由于之前i循环中判断得知a[i+1]一定大于a[i]
    //因此，将min_larger初始化为i+1
    while (j < size &&  a[j] > a[i])
    {
        if (a[min_larger] > a[j])
            min_larger = j;
        j++;
    }

    swap(a[i], a[min_larger]);                   
    string::iterator kb = a.begin() + i + 1;
    string::iterator ke = a.end();
    reverse(kb, ke);
    
    
       return false;

}




//计算下一个排列数
bool next_k_permutation(string &a,int length,string &result)
{
    reverse(a.begin() + length, a.end());
    
    if (!next_permutation(a))
    {
        string ss(a, 0, length);
        result = ss;
        return true;
    }
      return false;
    
}

void count_all(string&a)
{    
    int length = a.size();
    
    int count = 0;

    while (length >= 4)
    {    
        sort(a.begin(), a.end());
        count++;
        string result;
        while (next_k_permutation(a, length,result))
        {
            if (is_legal(result))
            {
                count++;
                
            }
        }

        --length;
        
        
    }

    cout << count << endl;
}

int main()
{
    string s = "123456789";
    
    count_all(s);

    
    system("pause");
    return 0;
}