UVa 11916 (discrete logarithm) emoogle Grid

Because the topic requires two different colors to be adjacent to the same column, there is no effect between columns and columns, which can be dyed on a column-by-row basis.

If a lattice above the adjacent lattice, has been dyed the lattice when dyed, a total of k-1 selected.

Conversely, if a lattice is in the first column, or the adjacent lattice above is a lattice that cannot be dyed, then there is a total of k selected.

Although, the number of rows in the matrix varies, but at least the maximum value of m for all rows that cannot be stained by the grid.

Test the number of scenarios (mod 100000007) for M-and m+1-lines respectively for R

Otherwise, the number of per-line schemes is multiplied by P = (k-1) n, assuming that the number of m+1 line schemes previously calculated is CNT

The following task is to solve px * cnt = r (mod mod), on both sides multiply CNT inverse, get px = R * cnt-1 (mod mod)

Finally add the previous m+1 line, the answer is x + M + 1

Finally Spit groove I think the pit point =_=, see that mold take for granted that is 1e9+7, but the last kind of example tune for a long time, and later only found the model is 1e8+7

1#include <cstdio>2#include <map>3#include <Set>4#include <cmath>5#include <algorithm>6 using namespacestd;7 #defineMP Make_pair8 #defineINS Insert9typedefLong LongLL;Ten  One Const intMOD =100000007; A Const intMAXB = -+Ten; - intN, M, K, B, R, X[maxb], Y[MAXB], Fir_row; - Set<pair<int,int> >Set; the  - intMul_mod (intAintb) -{return(LL) A * b%MOD;} -  + intPow_mod (intA, LL N) - { +     intAns =1,Base=A; A      while(n) at     { -         if(N &1) ans = mul_mod (ans,Base); -         Base= Mul_mod (Base,Base); -N >>=1; -     } -     returnans; in } -  to intInvinta) +{returnPow_mod (A, mod-2); } -  the intLog_mod (intAintb) *{//a^x=b (mod mod) $     intM, V, e =1, I;Panax Notoginsengm = (int) sqrt (MOD +0.5); -v =INV (Pow_mod (A, M)); themap<int,int>x; +x[1] =0; A      for(i =1; I < m; i++) the     { +E =Mul_mod (E, a); -         if(E = = b)returni; $         if(!x.count (e)) x[e] =i; $     } -      for(i =0; I < m; i++) -     { the         if(X.count (b))returnI*m +X[b]; -b =Mul_mod (b, v);Wuyi     } the     return-1; - } Wu  - intSolve () About { $     intc =0;//number of squares of k colors applied to the first m line -      for(inti =0; I < b; i++) -         if(X[i]! = M &&! Set.count (MP (x[i) +1, Y[i]))) C + +; -c + = n-Fir_row; A     intCNT = Mul_mod (Pow_mod (k, c), Pow_mod-K1, (LL) M*n-b-c)); +     if(cnt = = r)returnm; the  -c =0;//section m+1 the number of squares coated with K colors $      for(inti =0; I < b; i++)if(X[i] = = m) C + +; theCNT = Mul_mod (Mul_mod (CNT, Pow_mod (k, c)), Pow_mod (K-1, N-c)); the     if(cnt = = r)returnM +1; the  the     intp = Pow_mod (K-1, n); -     intv =INV (CNT); in     returnLog_mod (P, Mul_mod (R, v)) + M +1; the } the  About intMain () the { the     //freopen ("In.txt", "R", stdin); the  +     intT; -scanf"%d", &T); the      for(intKase =1; Kase <= T; kase++)Bayi     { thescanf"%d%d%d%d", &n, &k, &b, &R); the set.clear (); -m =1; Fir_row =0; -          for(inti =0; I < b; i++) the         { thescanf"%d%d", &x[i], &y[i]); the Set.ins (MP (X[i], y[i])); the             if(X[i] > m) m =X[i]; -             if(X[i] = =1) fir_row++;//the number of the first line that cannot be painted the         } theprintf"Case %d:%d\n", Kase, Solve ()); the     }94  the     return 0; the}

code June

UVa 11916 (discrete logarithm) emoogle Grid