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Recurrences UVA - 10870 矩阵快速幂

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题目链接:点我

题目

题意:

如题给出的公式,让你求 f(n) mod m的值;

思路:

矩阵快速幂;
根据题意构造矩阵为:

(a1a2a3a4....ad1000....00100....0..............................010)\begin{pmatrix} a_1 & a_2 & a_3 & a_4 &.... & a_d\\ 1 & 0 & 0 & 0 &.... &0\\ 0&1&0&0&....&0\\ ...&....&...&....&....&...\\ ..&...&....&0&1&0\\ \end{pmatrix} 

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

typedef long long LL;
int n, mod,d;

struct mat{
    LL a[20][20];
    mat(){memset(a, 0,sizeof(a)); }
    mat operator *(const mat q){
        mat c;
        for(int i = 1; i <= d; ++i)
            for(int k = 1; k <= d;++k)
            if(a[i][k])
        for(int  j = 1; j<= d; ++j){
            c.a[i][j] += a[i][k] * q.a[k][j];
            if(c.a[i][j] >= mod) c.a[i][j] %= mod;
        }return c;
    }
};

mat qpow(mat x, int n){
    mat ans;
    for(int i = 1; i <= 15; ++i)
        ans.a[i][i] = 1;
    while(n){
        if(n&1) ans =ans * x;
        x =x *x ;
        n >>= 1;
    }return ans;
}

int main(){
    int f[20];
    while(scanf("%d %d %d", &d, &n ,&mod) != EOF&&(d||mod||n)){
        mat ans;
        for(int i = 1; i <= d; ++i)
            scanf("%d", &ans.a[1][i]);
        for(int i = d; i > 0; --i)
            scanf("%d", &f[i]);
        if(n <= d){
            printf("%d\n",f[n] % mod);
            continue;
        }for(int i = 2; i <= d; ++i)
            ans.a[i][i-1] = 1;
        ans = qpow(ans,n-d);
        int sum = 0;
        for(int i = 1; i <= d; ++i)
            sum = (sum + f[i] * ans.a[1][i]) % mod;
        printf("%d\n",sum);
    }return 0;
}