A simple recursive problem (fast power + matrix multiplication optimization + scrolling array)

Question F: A simple recursive problem time limit: 1 Sec memory limit: MB
Submitted by: 546 resolution: 48
Submitted State [Discussion Version] The topic description exists as follows: F (n+1) =a1*f (n) +a2*f (n-1) +...+an*f (1) Finding the result input of the value of the K term to 1000000007 modulo

Single set of test data

First line input two integers n, K (1<=n<=100,n<k<=10000000000)

The second line enters n integers f (1) F (2) ... F (N)

Line three enter n integers A1 A2 ... An

Output

Output an integer

Sample input

2   to   4

Sample output

10


Today to do this problem, found that this problem is the card of the time of the divine problem ... Let me learn a lot.
-The first one is me as usual on the quick power of the template, return a matrix structure, but I found that even run can not run = =, after the discovery is due to the matrix structure in the body opened a 200*200 two-dimensional array, the two functions can not return such a large space structure,
I began to think that metaphysics, has been changed, changed for 1 hours, then the constant changed to 100, did not expect to run, unexpectedly is the reason, and learn to one!

-The second one is I changed after the AC not, has been reminded of the time overrun, and then the group Dalao teach me new posture: optimized matrix multiplication and scrolling array.

Matrix mult (Matrix A,matrix B,intN)    {Matrix temp;  for(intI=0; i<n; i++)    {         for(intk=0; k<n; k++)        {            if(A.v[i][k] = =0)Continue;  for(intj=0; j<n; J + +) {Temp.v[i][j]= (temp.v[i][j]+ (a.v[i][k]*b.v[k][j])%mod)%MoD; }        }    }    returntemp;}

Previous:

 matrix mult (Matrix A,matrix b,int n) {matrix temp;  for  (int  i= 0 ; I<n ; i++  for  (int  j=0 ; J<n ; j++" { for  (int  K=0 ; K<n ; K++) {            TEMP.V[I][J]  = (temp.v[i][j]+ (a.v[i][k]*b.v[k][j])%mod)%MOD; }}}  return   temp;}  

This technique is very useful! Especially when the matrix is high-order and sparse.

The optimization of the scrolling array is really metaphysical ... I don't know why it's so fast.

The coefficient matrix for this problem is:

A1 A2 ... an1   0    00   1   ...  0          .... 0 0     1 0

AC Code:

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<map>#include<vector>using namespaceStd;typedefLong LongLL;ConstLL mod =1000000007;Const intN =102; LL F[n];structmatrix{LL V[n][n]; Matrix () {memset (V,0,sizeof(v)); }} m[2]; Matrix mult (Matrix A,matrix B,intN)    {Matrix temp;  for(intI=0; i<n; i++)    {         for(intk=0; k<n; k++)        {            if(A.v[i][k] = =0)Continue;  for(intj=0; j<n; J + +) {Temp.v[i][j]= (temp.v[i][j]+ (a.v[i][k]*b.v[k][j])%mod)%MoD; }        }    }    returntemp;}voidpow_mod (LL t,ll N) { for(intI=0; i<n; i++) {m[0].v[i][i] =1; }     while(t) {if(t&1) m[0] = mult (m[0],m[1],n); m[1] = mult (m[1],m[1],n); T>>=1; }}intMain () {intN;    LL K; scanf ("%d%lld",&n,&k);  for(intI=0; i<n; i++) {scanf ("%lld", &f[n-1-i]); }     for(intI=0; i<n; i++) {scanf ("%lld", &m[1].v[0][i]); }     for(intI=1; i<n; i++) {m[1].v[i][i-1] =1; } k= kN;    Pow_mod (K,n); LL ans=0;  for(intI=0; i<n; i++) {ans= (ans+m[0].v[0][I]*F[I]%MOD)%MoD; } printf ("%lld\n", ans); return 0;}

A simple recursive problem (fast power + matrix multiplication optimization + scrolling array)