UVA 10870 - Recurrences 矩阵快速幂

矩阵快速幂模板

套模板即可

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;

typedef long long ll;
const int N = 20;
int msize, Mod;

struct Mat
{
    ll mat[N][N];
};

Mat operator *(Mat a, Mat b)
{
    Mat c;
    memset(c.mat, 0, sizeof(c.mat));
    for(int k = 0; k < msize; ++k)
        for(int i = 0; i < msize; ++i)
            if(a.mat[i][k])
                for(int j = 0; j < msize; ++j)
                    if(b.mat[k][j])
                        c.mat[i][j] = (c.mat[i][j] +a.mat[i][k] * b.mat[k][j])%Mod;
    return c;
}

Mat operator ^(Mat a, int k)
{
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    for(int i = 0; i < msize; ++i)
        c.mat[i][i]=1;
    for(; k; k >>= 1)
    {
        if(k&1) c = c*a;
        a = a*a;
    }
    return c;
}

int main()
{
//    freopen("in.txt","r",stdin);
    int n;
    int a[20] ,f[20];
    while(~scanf("%d%d%d", &msize, &n, &Mod))
    {
        if(msize == 0) break;
        for(int i = 0; i < msize; i++)
            scanf("%d", &a[i]);
        for(int i = msize-1; i >= 0; i--)
            scanf("%d", &f[i]);
        Mat A;
        memset(A.mat,0,sizeof(A.mat));
        for(int i = 0; i < msize; i++)
            A.mat[0][i] = a[i];
        for(int i = 1; i < msize; i++)
            A.mat[i][i-1] = 1;
        A = A^(n - msize);
        ll ans = 0;
        for(int i=0; i < msize; i++)
            ans = (ans + A.mat[0][i]*f[i]) % Mod;
        printf("%lld\n", ans);
    }
    return 0;
}