Kuangbin OJ 1217--Operations on Grids (YY) (Good question)

1217:operations on Grids time limit: 2 Sec Memory Limit: MB
Submit: Solved: 20
[Submit] [Status] [Web Board] Description

You have a 9-digit string, and now you have to fill in the 3x3 lattice with every bit of the number. If the number is 123456789, then fill in the 3x3 lattice to get: 123

456 789 Now you can do four things with this 3x3 lattice:

now give you this 9-bit number string, you need to do a  Left rotation operation,   b  Right rotation operation,   c    horizontal rollover operation,   d  The vertical flip operation, the sequence between these operations you decide.   asked how many different 3x3 squares can be obtained. Two 3x3 lattices are considered different when and only if there is at least one position in the   lattice.

Input

Enter the first line is an integerT t    each group of data has two rows. The first behavior is a 9-bit number string. The second line consists of 4 integers   a ,  b ,  C   d  (0 <=  a ,   b ,  C ,  D   <= 2)

Output

For each set of test data, the output of an integer indicates how many different 3x3 squares can eventually be obtained.

Sample Input

10000000001 2 1 2

Sample Output

1

HINT


train of thought: Because A,b,c,d is small, direct Dfs also can, but more yy also quite happy
yy After the conclusion:

C,d's only role is to:1. If C,d is 0, then the solution is only one case2.c,d One of them is not 0 .when the turn face, a actually turn right, b actually turn left, so, when c,d a not for 0 can let a, b contribute to each other, how much contribution (not more than a A, a self) only depends on and when to turn the face

when the figure is up or down symmetrical then the solution is 1
when the center of the graph is symmetrical then left 2 times and 4 times the same
In These three cases:

1312 KB 0 msc++ 1491 B 2015-03-21 23:04:38#include<cstdio> #include <iostream> #include <algorithm>#    Include<cstring>using namespace Std;char num[15];bool vis[5];int main () {int T;    scanf ("%d", &t);        while (t--) {int ans=0;        BOOL flag=0;        memset (vis,0,sizeof (VIS));        scanf ("%s", num+1);        int a,b,c,d;        scanf ("%d%d%d%d", &a,&b,&c,&d); if (num[1]==num[3]&&num[7]&&num[9]&&num[3]==num[7]&&num[2]==num[4]&&num            [4]==num[6]&&num[6]==num[8]) {printf ("1\n");        Continue            } if (c==0&&d==0) {printf ("1\n");        Continue        } if (num[4]==num[6]&&num[2]==num[8]&&num[1]==num[9]&&num[3]==num[7]) flag=1;        int s=a-b;            if (!vis[(s+4)%4]) {ans++;            vis[(s+4)%4] = true;        if (flag) vis[(s+2)%4] =true; } for (int i=1;i<=b;i++) {s+=2;            if (!vis[(s+4)%4]) ans++;            vis[(s+4)%4] = true;        if (flag) vis[(s+2)%4]=true;        } s=a-b;            for (int i=1;i<=a;i++) {s-=2;            if (!vis[(s+8)%4]) ans++;            vis[(s+8)%4]=true;        if (flag) vis[(s+10)%4]=true;    } printf ("%d\n", ans); } return 0;}

Kuangbin OJ 1217--Operations on Grids (YY) (Good question)