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Description
Pattern lock security is generally used in Android handsets instead of a password. The pattern lock can be set by joining points on a 3 × 3 matrix in a chosen order. The points of the matrix are registered in a numbered order starting with 1 in the upper left corner and ending with 9 in the bottom right corner.
A valid pattern has the following properties:
Now you are given n active points, you need to find the number of valid pattern locks formed from those active points.
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
The first line contains an integer n (3 ≤ n ≤ 9), indicating the number of active points. The second line contains n distinct integers a1, a2, … an (1 ≤ ai ≤ 9) which denotes the identifier of the active points.
Output
For each test case, print a line containing an integer m, indicating the number of valid pattern lock.
In the next m lines, each contains n integers, indicating an valid pattern lock sequence. The m sequences should be listed in lexicographical order.
Sample Input
1
3
1 2 3
Sample Output
4
1 2 3
2 1 3
2 3 1
3 2 1
被这道题目坑哭了快。。。
给出几个数字，判断几种手机设置锁的方式，并输出，，做起来很简单，，只是，，一开始因用cout超时，折磨很久，又因为没有
输出，一直WA 好苦。。。。
# include <iostream> # include <cstdio> # include <cstring> using namespace std; int pl[362882][10]; int num[362882][10]; int h = -1; void Perm(int list[], int k, int m) { if(k == m) { h++; for(int i = 0; i <= m; i++) pl[h][i] = list[i]; } else for(int j = k; j <= m; j++) { swap(list[k], list[j]); Perm(list, k + 1, m); swap(list[k], list[j]); } } int map[20] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; int is[20]; bool pan(int x, int y) { if(x == 1) { if(y == 3 && is[2] == 0) return false; if(y == 7 && is[4] == 0) return false; if(y == 9 && is[5] == 0) return false; } if(x == 2) { if(y == 8 && is[5] == 0) return false; } if(x == 3) { if(y == 1 && is[2] == 0) return false; if(y == 7 && is[5] == 0) return false; if(y == 9 && is[6] == 0) return false; } if(x == 4) { if(y == 6 && is[5] == 0) return false; } if(x == 6) { if(y == 4 && is[5] == 0) return false; } if(x == 7) { if(y == 1 && is[4] == 0) return false; if(y == 9 && is[8] == 0) return false; if(y == 3 && is[5] == 0) return false; } if(x == 8) { if(y == 2 && is[5] == 0) return false; } if(x == 9) { if(y == 1 && is[5] == 0) return false; if(y == 3 && is[6] == 0) return false; if(y == 7 && is[8] == 0) return false; } return true; } bool f(int list[], int n) { memset(is, 0, sizeof(is)); for(int i = 0; i < n - 1; i++) { is[list[i]] = 1; // 判断到下一个是否可以 if(!pan(list[i], list[i + 1])) return false; } return true; } int jie(int n) { int sum = 1; while(n) { sum *= n; n--; } return sum; } int main() { int t; scanf("%d", &t); while(t--) { memset(pl, 0, sizeof(pl)); int a[12]; int n; scanf("%d", &n); for(int i = 0; i < n; i++) scanf("%d", &a[i]); h = -1; Perm(a, 0, n - 1); int jiecheng = jie(n); int jishu = -1; for(int i = 0; i < jiecheng; i++) { if(f(pl[i], n)) // 符合条件 { jishu++; for(int j = 0; j < n; j++) { num[jishu][j] = pl[i][j]; //if(j == 0) //cout << pl[i][j]; //else //cout << " " << pl[i][j]; } //cout << endl; } } printf("%d\n", jishu + 1);//cout << jishu + 1 << endl; for(int i = 0; i < jishu + 1; i++) { for(int j = 0; j < n; j++) { if(j == 0) printf("%d", num[i][j]);//cout << num[i][j]; else printf(" %d", num[i][j]);//cout << " " << num[i][j]; } printf("\n");//cout << endl; } } return 0; }
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原文地址：http://www.cnblogs.com/lyf-acm/p/5453503.html
