POJ 2118 矩阵乘法

题意：

_an=Σ_{1<=i<=k   （}a_n-i*b_i） mod 10000

 

分析：

典型的矩阵乘法解线性递推式~

无限YM MATRIX67神犇~

 

http://www.matrix67.com/blog/archives/276/comment-page-1#comment-223435

 

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>

#define SIZE 110
#define mod 10000

using namespace std;

int n,m,a[SIZE],b[SIZE];

struct MT
{
    int x,y;
    int mt[SIZE][SIZE];
}ans,def;

inline MT operator *(MT a,MT b)
{
    MT c;
    memset(c.mt,0,sizeof c.mt);
    c.x=a.x; c.y=b.y;
    for(int i=1;i<=a.x;i++)
        for(int j=1;j<=b.y;j++)
            for(int k=1;k<=b.x;k++)
                c.mt[i][j]=(c.mt[i][j]+(a.mt[i][k]%mod)*(b.mt[k][j]%mod))%mod;
    return c;
}

bool read()
{
    for(int i=0;i<n;i++) scanf("%d",&a[i]);
    for(int i=0;i<n;i++) scanf("%d",&b[i]);
    scanf("%d",&m);
    if(m<=n-1) {printf("%d\n",a[m]);return false;}
    m=m-(n-1);
    for(int i=1;i<=n;i++) ans.mt[i][1]=a[i-1];
    ans.x=n; ans.y=1;
    memset(def.mt,0,sizeof def.mt);
    for(int i=1;i<n;i++) def.mt[i][i+1]=1;
    for(int i=1;i<=n;i++) def.mt[n][i]=b[n-i];
    def.x=def.y=n;
    return true;
}

void go()
{
    while(m)
    {
        if(m&1) ans=def*ans;
        def=def*def;
        m>>=1;
    }
    printf("%d\n",ans.mt[n][1]);
}

int main()
{
    while(scanf("%d",&n),n)
    {
        if(!read()) continue;
        else go();
    }
    return 0;
}