UVA 10870 - Recurrences 矩阵快速幂

                                      Recurrences

Consider recurrent functions of the following form:
f(n) = a1f(n − 1) + a2f(n − 2) + a3f(n − 3) + . . . + adf(n − d), for n > d,
where a1, a2, . . . , ad are 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, . . . , ad, 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, . . . , ad. The second set
gives values of f(1), f(2), . . . , f(d).
You can assume that: 1 ≤ d ≤ 15, 1 ≤ n ≤ 2
31 − 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
1 1 100
2
1
2 10 100
1 1
1 1
3 2147483647 12345
12345678 0 12345
1 2 3
0 0 0
Sample Output
1
55
423

 

题意：

给你 ：递推关系f(n)=a1*f(n-1) + a2*f(n-2) + .. ad*f(n-d)，计算f(n)%m

题解：

n太大，所以矩阵快速幂咯

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long ll;
const int N = 50;
ll m,d,f[N];
struct Matrix {
    ll mat[20][20];
};
Matrix multi (Matrix a, Matrix b) {
    Matrix ans;
    memset(ans.mat,0,sizeof(ans.mat));
    for(int i = 1; i <= d; i++) {
        for(int j = 1; j <= d; j++) {
            for(int k = 1; k <= d; k++)
                ans.mat[i][j] += a.mat[i][k] * b.mat[k][j],ans.mat[i][j] %= m;
        }
    }
    return ans;
}
ll powss(ll n,Matrix ans) {
    Matrix p,t;memset(t.mat,0,sizeof(t.mat));  
    for(int i = 1; i <= d; i++) t.mat[i][i] = 1;
    while(n) {
        if(n&1) t = multi(t,ans);
        n>>=1;
        ans = multi(ans,ans);
    }
     ll tmp=0;
    for(int i=1;i<=d;i++)
        tmp=(tmp+t.mat[1][i]*f[d - i + 1])%m;
    return tmp;
}
int main() {
    ll n,a[N];
    while(~scanf("%lld%lld%lld",&d,&n,&m)) {
        if(d == 0 && m == 0 &&n == 0) break;
        for(int i = 1; i <= d; i++) scanf("%lld",&a[i]);
        for(int i = 1; i <= d; i++) scanf("%lld",&f[i]);
        if(n <= d) {cout<<f[n]<<endl;continue;}
        Matrix ans;memset(ans.mat,0,sizeof(ans.mat));
        for(int i = 1; i <= d; i++) ans.mat[i][i-1] = 1;
        for(int i = 1; i <= d; i++) ans.mat[1][i] = a[i];
        printf("%lld\n",powss(n - d,ans));
    }
    return 0;
}