Sgu 197 nice patterns strike back | zoj 2317 nice patterns strike back

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Question recognition:After doing this, I learned a lot. First, I learned about the state compression push formula.

Solution:

Constructor matrix:B [I] [J] (I, j is in the state, 0 <= I, j <= (1 <m)-1 ), if B [I] [J] is equal to 1, it indicates that status I is compatible with status J (I, j can be in two adjacent rows). If it is 0, it indicates that it cannot be adjacent, there is also a matrix of the original matrix A, (1 <m)-1 rows, and 1 columns, representing the number of solutions in various states in the first row, in this way, B * A gets the number of solutions in various states in the second line, so ~ > A ^ (n-1) * B obtains the number of solutions in the nth row. The final matrix is the total number of solutions. Because the index is relatively large, so ~> You need to simulate the division of 2 for a large number.

Code:

# Include <iostream> # include <sstream> # include <map> # include <cmath> # include <fstream> # include <queue> # include <vector> # include <sstream> # include <cstring> # include <cstdio> # include <stack> # include <bitset> # include <ctime> # include <string> # include <cctype> # include <iomanip> # include <algorithm> using namespace STD; # define int long int # define L (x) (x * 2) # define R (x) (x * 2 + 1) const int INF = 0x3f3f3f; const do Uble ESP = 0.0000000001; const double Pi = ACOs (-1.0); const int my = 1400 + 5; const int MX = 1005 + 5; int S, M; int d [MX]; int MOD; struct STR {int N, d [MX]; STR () {n = 0; memset (D, 0, sizeof (d);} STR operator-(int x) {STR s; S. N = N; For (INT I = n-1; I> = 0; -- I) if (d [I]) {d [I] --; for (Int J = I + 1; j <n; ++ J) d [J] = 9; break;} If (d [0] = 0) {for (INT I = 1; I <n; ++ I) s. d [I-1] = d [I]; S. N --;} else for (INT I = 0; I <n; ++ I) s. d [I] = d [I]; return s;} STR operator> (INT X) {STR s; int A [MX], num = 0, sum = 0; for (INT I = 0; I <n; ++ I) {sum = sum * 10 + d [I]; If (sum> = 2) {A [num ++] = sum/2; sum = sum % 2;} else if (Num) A [num ++] = 0 ;} for (INT I = 0; I <num; ++ I) s. d [I] = A [I]; S. N = num; return s;} bool operator & (int x) {If (d [n-1] & 1) Return True; return false ;}}; struct m {int P [33] [33]; m () {memset (p, 0, sizeof (p ));} void Init () {for (INT I = 0; I <s; ++ I) P [I] [I] = 1 ;}}; bool judge (INT X, int y) {for (INT I = 0; I <m-1; ++ I) if (X & (1 <I )) & (X & (1 <(I + 1) & (Y & (1 <I )) & (Y & (1 <(I + 1) | (! (X & (1 <I ))&&! (X & (1 <(I + 1 )))&&! (Y & (1 <I ))&&! (Y & (1 <(I + 1) return false; return true;} m pre () // constructor {M C; for (INT I = 0; I <s; ++ I) for (Int J = I; j <s; ++ J) if (Judge (I, j )) c. P [I] [J] = C. P [J] [I] = 1; return C;} m operator * (const M & A, const M & B) // matrix multiplication {M C; for (INT I = 0; I <s; ++ I) for (int K = 0; k <s; ++ K) if (. P [I] [k]) for (Int J = 0; j <s; ++ J) C. P [I] [J] = (C. P [I] [J] +. P [I] [k] * B. P [k] [J]) % MOD; return C;} m POW (m a, str k) // fast power {m B; B. init (); While (K. n) {If (K & 1) B = B * A; A = A * A; k = k> 1;} return B;} int main () {char STR [MX]; while (CIN> STR> m> mod) {S = (1 <m); // The total status of str s; for (INT I = 0; I <strlen (STR); ++ I) s. d [I] = STR [I]-'0'; S. N = strlen (STR); s = s-1; m c = pre (); m B = POW (C, S); int ans = 0; for (INT I = 0; I <s; ++ I) for (Int J = 0; j <s; ++ J) ans = (ANS + B. P [I] [J]) % MOD; cout <ans <Endl;} return 0 ;}

Sgu 197 nice patterns strike back | zoj 2317 nice patterns strike back