"Bzoj" "3240" "NOI2013" Matrix game

Decimal fast Power + matrix multiplication + constant optimization

I heard that this problem can also be forced to calculate the recurrence of the type ... Then take multiplication to calculate it ...

However Konjac Konjac opted for a more violent approach = =

We find that this recursive process is linear, so we can use matrix multiplication to represent, $x =a*x+b$ such a recursive we can say: $$\begin{bmatrix} x& 1 \end{bmatrix} * \begin{bmatrix} A & 0 \ b& 1 \end{bmatrix} $$

Then we can make $s_1$ express xa+b, $s _2$ xc+d, then we have $ $ans =v * (({s_1}^{n-1}*s_2) ^{m-1} * {s_1}^{n-1}) $$

But I gave Tle a direct count ...

　　

Here's a look at constant optimization:

We noticed that when matrix multiplication was: $$\begin{bmatrix} a& 0 \ b& 1 \end{bmatrix} * \begin{bmatrix} c& 0 \ d& 1 \end{bmatrix} = \begin{bmatrix} a*c& 0 \ a*d+b& 1 \end{bmatrix}$$

That is to say: The second column of 0 and 1 is not moving ... Then we can optimize the matrix multiplication process of most $o (n^3) $ to $o (n^2) $.

Here we ${s_1}^{n-1}$ two times, then we can use an intermediate variable to save first, can reduce the first operation (after all, the main part of the algorithm is to calculate the power)

1 /**************************************************************2 problem:32403 User:tunix4 language:c++5 result:accepted6 time:7980 Ms7 memory:3232 KB8 ****************************************************************/9  Ten //Bzoj 3240 One#include <vector> A#include <cstdio> -#include <cstring> -#include <cstdlib> the#include <iostream> -#include <algorithm> - #defineRep (i,n) for (int i=0;i<n;++i) - #defineF (i,j,n) for (int i=j;i<=n;++i) + #defineD (i,j,n) for (int i=j;i>=n;--i) - #definePB Push_back + using namespacestd; AInlineintGetint () { at     intv=0, sign=1;CharCh=GetChar (); -      while(ch<'0'|| Ch>'9'){if(ch=='-') sign=-1; Ch=GetChar ();} -      while(ch>='0'&&ch<='9') {v=v*Ten+ch-'0'; Ch=GetChar ();} -     returnv*Sign ; - } - Const intn=1e6+Ten, inf=~0u>>2, p=1e9+7; intypedefLong LongLL; - /******************tamplate*********************/ to   + structmatrix{ -     intv[2][2]; theMatrix (intx=0) {F (I,0,1) F (J,0,1)if(i==j) v[i][j]=x;Elsev[i][j]=0;} *     int*operator[] (intx) {returnv[x];} $ }s1,s2,v;Panax NotoginsengInline Matrixoperator*(Matrix A,matrix b) { - Matrix C; the     if(a[0][1]==0&& a[1][1]==1&& b[0][1]==0&& b[1][1]==1){ +c[0][0]= (LL) a[0][0]*b[0][0]%P; Ac[0][1]=0; thec[1][0]= (LL) a[0][0]*b[1][0]+ (LL) a[1][0])%P; +c[1][1]=1; -         returnC; $     } $F (k,0,1) F (I,0,1) F (J,0,1) -c[i][j]= (LL) c[i][j]+ (LL) a[i][k]*b[k][j]%p)%P; -     returnC; the } -The inline matrix Pow (matrix A,intb) {WuyiMatrix C (1); theF (I,1, b) c=c*A; -     returnC; Wu } -Inline matrix Power (matrix A,Char*s) { AboutMatrix R (1);intL=strlen (s); $D (i,l-1,0){ -         if(s[i]-'0') R=r*pow (a,s[i]-'0'); -A=pow (A,Ten); -     } A     returnR; + } the CharN[n],m[n]; - intMain () { $ #ifndef Online_judge theFreopen ("3240.in","R", stdin); theFreopen ("3240.out","W", stdout); the #endif thescanf"%s", n); scanf"%s", m); -     intL1=strlen (N)-1; in      while(n[l1]=='0') n[l1--]='9'; then[l1]--; theL1=strlen (M)-1; About      while(m[l1]=='0') m[l1--]='9'; them[l1]--; the //printf ("%s%s\n", n,m); the     inta,b,c,d; +A=getint (); B=getint (); C=getint (); D=getint (); -v[0][0]=v[0][1]=1; v[1][0]=v[1][1]=0; thes1[0][0]=a; s1[0][1]=0; s1[1][0]=b; s1[1][1]=1;Bayis2[0][0]=c; s2[0][1]=0; s2[1][0]=d; s2[1][1]=1; theMatrix s3=Power (s1,m); thev=v* (Power (s3*s2,n) *S3); -printf"%d\n", v[0][0]); -     return 0; the}

View Code 3240: [Noi2013] Matrix game time limit:10 Sec Memory limit:256 MB
submit:890 solved:390
[Submit] [Status] [Discuss] Description

Tingting is a love matrix of the children, one day she wants to use a computer to generate a huge n-row m-column matrix (you do not have to worry about how she stored). The matrix that she generates satisfies a magical nature: if you use F[i][j] to represent the element in column J of the Matrix, then F[i][j] satisfies the following recursion:

F[1][1]=1
F[i,j]=a*f[i][j-1]+b (J!=1)
F[i,1]=c*f[i-1][m]+d (I!=1)
The a,b,c,d in a recursive style is a given constant.

Now tingting want to know f[n][m] The value is how much, please help her. Since the end result can be large, you only need to output F[N][M] divided by 1,000,000,007 remainder.

Input

A row has six integer n,m,a,b,c,d. The meaning title said

Output

Contains an integer that represents the remainder of f[n][m] divided by 1,000,000,007

Sample Input3 4 1 3 2 6Sample Output -HINT

The matrix in the sample is:

1 4 7 10

26 29 32 35

76 79 82 85

1<=n,m<=10^1000 000,a<=a,b,c,d<=10^9

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"Bzoj" "3240" "NOI2013" Matrix game