"Solving" matrix Bzoj 4128 matrices inverse discrete logarithm step-by-step algorithm

Portal: http://www.lydsy.com/JudgeOnline/problem.php?id=4128

A flood problem together

Using the STRIDE step algorithm, the operation of the number into the matrix is good.

Matrix inversion? So the basis of the line-generation algorithm I do not want to say, or self-baidu it

It is important to note that matrices do not have a commutative law, so do not confuse the sequence when calculating $b\cdot a^{-m}$.

Code:

1#include <cstring>2#include <cstdio>3#include <algorithm>4#include <map>5#include <cmath>6 7 using namespacestd;8 Const intMAXN = -;9 Ten intn, p; One  A structMatrix { -     intA[MAXN][MAXN]; -     int*operator[](intidx) { the         returnA[idx]; -     } -     Const int*operator[](intIDX)Const { -         returnA[idx]; +     } -     BOOL operator< (ConstMatrix &AMP;RHS)Const{//for map +          for(inti =0; I < n; ++i) A              for(intj =0; J < N; ++j) at                 if(A[i][j] < rhs[i][j])return true; -                 Else if(A[i][j] > Rhs[i][j])return false; -         return false; -     } -     voidUnit () {//fill in the N-Order unit matrix -Memset (A,0,sizeof(a)); in          for(inti =0; I < n; ++i) A[i][i] =1; -     } to     voidClear () {//Complete 0 +Memset (A,0,sizeof(a)); -     } the }; *Matrix A, B;//given A/b matrix in the title $ Panax Notoginseng intPow_mod (intAintb) {//integer Fast Power -     intAns =1; the      while(b) { +         if(b&1) ans = ans * a%p; AA = a * a%p; theb >>=1; +     } -     returnans; $ } $ intInvintA) {//Inverse integer negation -     returnPow_mod (a,p-2); - } the voidInput_matrix (Matrix &A) { -      for(inti =0; I < n; ++i)Wuyi          for(intj =0; J < N; ++j) { thescanf"%d", &a[i][j]); -A[I][J]%=p; Wu         } - } About voidMulConstMatrix &a,ConstMatrix &b, Matrix &c) {//matrix multiplication, results deposited in C $ Matrix tmp; Tmp.clear (); -      for(inti =0; I < n; ++i) -          for(intj =0; J < N; ++j) -              for(intK =0; K < n; ++k) ATMP[I][J] = (Tmp[i][j] + a[i][k] * B[k][j]% p)%p; +C =tmp; the } - voidGauss_jordan (inta[maxn][maxn<<1] ) {//The inverse of Gausio when eliminating Yuan $      for(inti =0; I < n; ++i) { the         intR; the          for(r = i; r < N; + +)R) the             if(A[r][i]) Break; the         if(r! =i) -              for(intj = i; J < (n<<1); ++j) in swap (A[r][j], a[i][j]); the         intIV =INV (a[i][i]); the          for(intj = i; J < (n<<1); ++j) AboutA[I][J] = a[i][j] * IV%p; the          for(intj =0; J < N; ++j) the             if(J! = i &&A[j][i]) { the                 intTMP = (P-a[j][i])%p; +                  for(intK = i; K < (n<<1); ++k) -A[j][k] = (A[j][k] + a[i][k] * tmp% p)%p; the             }Bayi     } the } the voidInvConstMatrix &a, Matrix &b) {//matrix inversion, results deposited b -     inttmp[maxn][maxn<<1] = {0}; -      for(inti =0; I < n; ++i) the          for(intj =0; J < N; ++j) theTMP[I][J] =A[i][j]; the      for(inti =0; I < n; ++i) Tmp[i][i+n] =1; the Gauss_jordan (TMP); -      for(inti =0; I < n; ++i) the          for(intj =0; J < N; ++j) theB[I][J] = tmp[i][j+n]; the }94  theMap<matrix,int>MP; the intBSGS (Matrix A, matrix B) {//Step -by-step algorithm the     intm = sqrt (p) +1;98 Matrix E; E.unit (); About      for(inti =0; I < m; ++i) { -         if(!mp.count (E)) mp[e] =i;101 Mul (e,a,e);102     }103 INV (e,e);104      for(inti =0; I < P; i + =m) { the         if(Mp.count (B))returni +Mp[b];106Mul (B,E,B);//Note Matrix multiplication order107     }108     return-1;109 } the 111 intMain () { thescanf"%d%d", &n, &p);113 Input_matrix (A), Input_matrix (B); theprintf"%d\n", Bsgs (A, b)); the     return 0; the}

"Solving" matrix Bzoj 4128 matrices inverse discrete logarithm step-by-step algorithm