King's Phone HDU - 5641（模拟）

最新推荐文章于 2020-11-11 18:53:01 发布

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本文介绍了一个手机图案解锁的有效性验证算法。该算法基于特定规则判断输入的图案密码是否有效，包括密码长度、不可重复经过同一节点及路径上的中间点不能被跳过等约束条件。

King's Phone

HDU - 5641

In a military parade, the King sees lots of new things, including an Andriod Phone. He becomes interested in the pattern lock screen.

The pattern interface is a 3×3 square lattice, the three points in the first line are labeled as 1,2,3, the three points in the second line are labeled as 4,5,6, and the three points in the last line are labeled as 7,8,9。The password itself is a sequence, representing the points in chronological sequence, but you should follow the following rules:

- The password contains at least four points.

- Once a point has been passed through. It can't be passed through again.

- The middle point on the path can't be skipped, unless it has been passed through( 3427 is valid, but 3724 is invalid).

His password has a length for a positive integer k(1≤k≤9), the password sequence is s1,s2...sk(0≤si<INT_MAX)

, he wants to know whether the password is valid. Then the King throws the problem to you.

Input The first line contains a number  T(0<T≤100000), the number of the testcases.

For each test case, there are only one line. the first first number  k，represent the length of the password, then k numbers, separated by a space, representing the password sequence s1,s2...sk. Output Output exactly T lines. For each test case, print `valid` if the password is valid, otherwise print `invalid` Sample Input

Sample Output

invalid
valid
valid

hint:
For test case #1：The path $1\rightarrow 3$ skipped the middle point $2$, so it's invalid.

For test case #2：The path $1\rightarrow 3$ doesn't skipped the middle point $2$, because the point 2 has been through, so it's valid.

For test case #2：The path $8\rightarrow 1 \rightarrow 6 \rightarrow 7$ doesn't have any the middle point $2$, so it's valid.

主要是是搞懂题目意思

手机按键1 2 3

4 5 6

7 8 9

三个条件

这里条件3的路径是指按键路径，比如一行123，这个路径如果有13，跳过了中间2，而2之前没出现过就算无效密码

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int vis[15];
int judge(int a,int b){
//    1 2 3
//    4 5 6
//    7 8 9
    if(a > b) swap(a,b);
    int c = b - a;//计算两个数的差
    if(c == 2 && (a == 1 || a == 4 || a == 7) && !vis[(a+b)/2]) return 0;//如果是每行，中间跳过了返回0
    if(c == 4 && a == 3 && !vis[(a+b)/2]) return 0;//如果是副对角线，中间跳过了返回0
    if((c == 6 || c == 8) && !vis[(a+b)/2]) return 0;//如果是每列，或者主对角线，中间跳过了，返回0
    return 1;
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int k,i,j;
        int valid = 1;
        int password[15];
        memset(vis,0,sizeof(vis));
        scanf("%d",&k);
        for(i = 0; i < k; i++){
            scanf("%d",&password[i]);
            if(password[i] < 1 || password[i] > 9) valid = 0;
        }
        if(k < 4 || k > 9 || !valid){
            printf("invalid\n");
            continue;
        }
        vis[password[0]] = 1;
        for(i = 1; i < k; i++){
            vis[password[i]]++;
            if(!judge(password[i-1],password[i]) || vis[password[i]] > 1) valid = 0;
        }
        if(valid) printf("valid\n");
        else printf("invalid\n");
    }
    return 0;
}