uva 10870 递推关系矩阵快速幂模

Recurrences Input: standard input Output: standard output

Consider recurrent functions of the following form:

f(n) = a₁ f(n - 1) + a₂ f(n - 2) + a₃ f(n - 3) + ... + a_d f(n - d), for n > d. a₁, a₂, ..., a_d - arbitrary constants.

A famous example is the Fibonacci sequence, defined as: f(1) = 1, f(2) = 1, f(n) = f(n - 1) + f(n - 2). Here d = 2, a₁ = 1, a₂ = 1.

Every such function is completely described by specifying d (which is called the order of recurrence), values of d coefficients: a₁, a₂, ..., a_d, and values of f(1), f(2), ..., f(d). You'll be given these numbers, and two integers n and m. Your program's job is to compute f(n) modulo m.

Input

Input file contains several test cases. Each test case begins with three integers: d, n, m, followed by two sets of d non-negative integers. The first set contains coefficients: a₁, a₂, ..., a_d. The second set gives values of f(1), f(2), ..., f(d).

You can assume that: 1 <= d <= 15, 1 <= n <= 2³¹ - 1, 1 <= m <= 46340. All numbers in the input will fit in signed 32-bit integer.

Input is terminated by line containing three zeroes instead of d, n, m. Two consecutive test cases are separated by a blank line.

Output

For each test case, print the value of f(n) (mod m) on a separate line. It must be a non-negative integer, less than m.

Sample Input                              Output for Sample Input

1 1 100 2 1   2 10 100 1 1 1 1   3 2147483647 12345 12345678 0 12345 1 2 3 0 0 0

1 55 423

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题目大意：f(n)=a1*f(n-1)+a2*f(n-2)+.....+ad*f（n-d）

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

#define Max 20
typedef long long LL;

struct Matrix
{
    LL a[Max][Max];
    int n;
};

Matrix Matrix_mult_mod(Matrix A,Matrix B,int m)
{
    int i,j,k;
    Matrix C;
    C.n=A.n;
    memset(C.a,,sizeof(C.a));
    for(i=;i<=A.n;i++)
    {
        for(j=;j<=A.n;j++)
        {
            for(k=;k<=A.n;k++)
            {
                C.a[i][j]=(C.a[i][j]+A.a[i][k]*B.a[k][j])%m;
            }
        }
    }
    return C;
}

Matrix Matrix_pow_mod(Matrix A,int n,int m)
{
    Matrix t;
    int i,j;
    t.n=A.n;
    memset(t.a,,sizeof(t.a));
    for(i=;i<=A.n;i++) t.a[i][i]=;
    for(i=;i<=A.n;i++)
    for(j=;j<=A.n;j++)
        A.a[i][j]%=m;
    while(n)
    {
        if(n&) t=Matrix_mult_mod(t,A,m);
        n>>=;
        A=Matrix_mult_mod(A,A,m);
    }
    return t;
}

void deal(int d,int n,int m)
{
    int i,j;
    LL dd[Max],dd1[Max];
    Matrix A;
    A.n=d;
    memset(A.a,,sizeof(A.a));
    for(i=,j=;j<=d;i++,j++) A.a[i][j]=;
    for(j=d,i=;i<=d;i++,j--) scanf("%ll",&A.a[d][j]);
    for(i=;i<=d;i++) scanf("%ll",dd+i);
    A=Matrix_pow_mod(A,n-d,m);
    for(i=;i<=d;i++)
    {
        dd1[i]=;
        for(j=;j<=d;j++)
            dd1[i]=(dd1[i]+A.a[i][j]*dd[j])%m;
    }
    printf("%ll\n",dd1[d]);
}

int main()
{
    int d,n,m;
    while(scanf("%d %d %d",&d,&n,&m),d+n+m)
        deal(d,n,m);
    return ;
}