ACM 矩阵乘法 SPOJ AMR10E Stocks Prediction

http://www.spoj.com/problems/AMR10E/

E - Stocks Prediction

Time Limit:8000MS     Memory Limit:KB     64bit IO Format:%lld & %llu

 SPOJ AMR10E

Description

The department store where my family goes shopping is trying to predict how much of each item they stock they will sell each month.  Keeping a large amount of inventory is costly, and running out of items is also not a good idea.  Since the manager asked for my help as a sales consultant, I decided to formulate a model for predicting each month's sales S of an item based on its sales during the previous R months.  After a lot of trial and error, I came up with such a model, where 
S(n) = a1*S(n-1) + a2*S(n-2) + ... + aR*S(n-R)  
where S(n) is the predicted sales for the nth month for n > R, and S(1) to S(R) are seed values. 
 
The store manager was pleased with my model's ability to help him in controlling his inventory.  
He asked me to list out every Kth month's sales, and give him the sum of the first N values from this list.  For example he wanted every Christmas month's sales summed up for the next 10 years (N=10 and K=12, month 1 being January), or every end-of-quarter month's sales for the next 2 years (N=2, K=3). 
 
Can you please help me write a program that does all the above? 
 
INPUT 
The first line of the input T, the number of test cases. Each test case consists of three lines. 
The first line of each test case contains N, R, K. 
The second line of each test case contains R integers denoting S(1), S(2), ..., S(R). 
The third line of each test case contains R integers denoting the coefficients a1, a2, ..., aR of the predictive model. 
 
OUTPUT 
For each test case, output the sum requested by the manager as given in the problem statement, modulo 1,000,000,007. 
 
CONSTRAINTS 
T <= 40 
1 <= N <= 1000000000 
1 <= R <= 8 
1 <= K <= 8 
0 <= All other input values < 1000000007 
 
SAMPLE INPUT 
2 
4 1 1 
1  
2  
3 2 3 
1 1  
1 1  
 
SAMPLE OUTPUT 
15 
44 
 
EXPLANATION 
In the first test case, it is given that S(1) = 1 and the relation is S(n)=2*S(n-1). The list asked by the store manager consists of all the terms of S since K is 1. Hence, the answer is just the sum of the first 4 terms of S. 
In the second test case, the sequence S is the fibonacci sequence which is: 1, 1, 2, 3, 5, 8, 13, 21, 34. The list consists of 2, 8, 34 which sum up to 44.

转移矩阵为a

先将分为每组k次，算a^k //a=quick(a,k);

再算(a^k)^k //a=quick(a,n);

<strong>Memory: </strong>2662 KB<strong>Time: </strong>170 MS<strong>Language: </strong>C++ (g++ 4.3.2)<strong>Result: </strong><span style="color:blue;">Accepted</span>

/* * Author: NICK WONG * Created Time:  7/30/2014 16:45:37 * File Name:  */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;#define out(x) cout<<#x<<": "<<x<<endlconst double eps(1e-8);const int maxn=10100;const long long maxint=-1u>>1;const long long maxlong=maxint*maxint;typedef long long lint;const lint mod=1000000007;lint ans;int n,r,k,p[20];lint s[200];struct matrix{    lint m[10][10];    matrix()    {        memset(m,0,sizeof(m));//注意要清零    }};matrix operator * (const matrix & a,const matrix & b){    matrix c;    for (int i=1; i<=9; i++)        for (int j=1; j<=9; j++)        {            c.m[i][j]=0;            for (int k=1; k<=9; k++)                c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%mod;        }    return c;}matrix quick(matrix base,int pow){    matrix a;    for (int i=1; i<=9; i++) a.m[i][i]=1;//单位阵，矩阵乘法时要用到    while (pow)    {        if (pow&1) a=a*base;        base=base*base;        pow>>=1;    }    //if (pow==0) return base;    return a;}            void init(){    memset(s,0,sizeof(s));    cin>>n>>r>>k;    for (int i=1; i<=r; i++) cin>>s[i];    for (int i=1; i<=r; i++) cin>>p[i];}void work(){    for (int i=r+1; i<=80; i++)    {        for (int j=1; j<=r; j++) s[i]=(s[i]+p[j]*s[i-j])%mod;    }    matrix a,wjj;    for (int i=1; i<=8; i++) wjj.m[1][i]=s[i];    for (int i=1; i<=7; i++) a.m[i+1][i]=1;    for (int i=1; i<=r; i++) a.m[8-i+1][8]=p[i];    a=quick(a,k);    a.m[k][9]=a.m[9][9]=1;    a=quick(a,n);    wjj=wjj*a;    ans=wjj.m[1][9];    cout<<ans<<endl;  }int main(){    int t;    cin >> t;    while (t--)    {           init();        work();    }    return 0;}