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2.37 kB
| // Let It Flow | |
| // Solution by Jacob Plachta | |
| using namespace std; | |
| template<typename T> T Abs(T x) { return(x<0 ? -x : x); } | |
| template<typename T> T Sqr(T x) { return(x*x); } | |
| string plural(string s) { return(Sz(s) && s[Sz(s)-1]=='x' ? s+"en" : s+"s"); } | |
| const int INF = (int)1e9; | |
| const LD EPS = 1e-12; | |
| const LD PI = acos(-1.0); | |
| bool Read(int &x) | |
| { | |
| char c,r=0,n=0; | |
| x=0; | |
| for(;;) | |
| { | |
| c=GETCHAR(); | |
| if ((c<0) && (!r)) | |
| return(0); | |
| if ((c=='-') && (!r)) | |
| n=1; | |
| else | |
| if ((c>='0') && (c<='9')) | |
| x=x*10+c-'0',r=1; | |
| else | |
| if (r) | |
| break; | |
| } | |
| if (n) | |
| x=-x; | |
| return(1); | |
| } | |
| int N; | |
| char G[3][1001]; | |
| int Solve() | |
| { | |
| int i,j,c,ans; | |
| // impossible? | |
| if (N%2) | |
| return(0); | |
| if ((G[0][0]=='#') || (G[2][N-1]=='#')) | |
| return(0); | |
| Fox(i,N) | |
| if (G[1][i]=='#') | |
| return(0); | |
| // count | |
| ans=1; | |
| for (i=1; i<N-1; i+=2) | |
| { | |
| c=0; | |
| for (j=0; j<=2; j+=2) | |
| if ((G[j][i]!='#') && (G[j][i+1]!='#')) | |
| c++; | |
| ans=ans*c%MOD; | |
| } | |
| return(ans); | |
| } | |
| int main() | |
| { | |
| if (DEBUG) | |
| freopen("in.txt","r",stdin); | |
| // vars | |
| int T,t; | |
| int i; | |
| // testcase loop | |
| Read(T); | |
| Fox1(t,T) | |
| { | |
| // input | |
| Read(N); | |
| Fox(i,3) | |
| scanf("%s",&G[i]); | |
| // output | |
| printf("Case #%d: %d\n",t,Solve()); | |
| } | |
| return(0); | |
| } |