Created
May 13, 2012 11:01
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number of patterns of android unlock screen
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#include <stdio.h> | |
int visited[9] = {0}; | |
int nsteps = 0; | |
/* alternate can_connect */ | |
int max(int x, int y) | |
{ | |
return x >= y? x: y; | |
} | |
int min(int x, int y) | |
{ | |
return x <= y? x: y; | |
} | |
int can_connect1(int i, int j) | |
{ | |
int x1 = i / 3, y1= i % 3; | |
int x2 = j / 3, y2= j % 3; | |
int dx = max(x1, x2) - min(x1, x2); | |
int dy = max(y1, y2) - min(y1, y2); | |
int mx = (x1+x2) / 2; | |
int my = (y1+y2) / 2; | |
int m = (i + j) / 2; | |
if (((dx * dy == 0 && dx + dy == 2) || | |
(dx == 2 && dy == 2)) && | |
!visited[m]) | |
return 0; | |
return 1; | |
} | |
/* */ | |
int can_connect(int i, int j) | |
{ | |
int x1 = i / 3, y1= i % 3; | |
int x2 = j / 3, y2= j % 3; | |
int mx = (x1+x2) / 2; | |
int my = (y1+y2) / 2; | |
int m = (i + j) / 2; | |
if (x1+x2 == mx * 2 && y1+y2 == my * 2 && !visited[m]) | |
return 0; | |
return 1; | |
} | |
int track(int p) | |
{ | |
int npath = 0; | |
int i; | |
visited[p] = 1; | |
nsteps++; | |
if (nsteps >= 4) | |
npath++; | |
for (i = 0; i < 9; i++) { | |
if (visited[i] || !can_connect(p, i)) | |
continue; | |
npath += track(i); | |
} | |
visited[p] = 0; | |
nsteps--; | |
return npath; | |
} | |
int solve() | |
{ | |
int ret = 0; | |
ret += 4 * track(0); | |
ret += 4 * track(1); | |
ret += track(4); | |
return ret; | |
} | |
int main() | |
{ | |
printf("%d\n", solve()); | |
return 0; | |
} |
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