POJ3070 Fibonacci（矩阵快速幂加速递推）【模板题】

题目链接：传送门

题目大意：

　　求斐波那契数列第n项F(n)。

　　（F(0) = 0, F(1) = 1, 0 ≤ n ≤ 10⁹）

思路：

　　用矩阵乘法加速递推。

算法竞赛进阶指南的模板：

#include <iostream>
#include <cstring>
 
using namespace std;
const int MOD = ;
 
void mul(int f[], int base[][]) {
    int c[];
    memset(c, , sizeof c);
    for (int j = ; j < ; j++) {
        for (int k = ; k < ; k++) {
            c[j] = (c[j] + 1LL * f[k] * base[k][j]) % MOD;
        }
    }
    memcpy(f, c, sizeof c);
}
void mulself(int base[][]) {
    int c[][];
    memset(c, , sizeof c);
    for (int i = ; i < ; i++)
        for (int j = ; j < ; j++)
            for (int k = ; k < ; k++)
                c[i][j] = (c[i][j] + 1LL*base[i][k]*base[k][j]) % MOD;
    memcpy(a, c, sizeof c);
}
 
int main()
{
    int n;
    while (cin >> n && n != -) {
        int f[] = {, };
        int base[][] = {{, }, {, }};
        for (; n; n >>= ) {
            if (n & ) mul(f, base);
            mulself(base);
        }
        cout << f[] << endl;
    }
    return ;
}

蒟蒻的模板：

#include <cstdio>
#include <iostream>
#include <cstring>
 
using namespace std;
typedef long long ll;
const int MOD = 1e4;
const int MAXN = ;
struct Matrix{
    int mat[MAXN][MAXN];
    Matrix operator * (Matrix const& b) const {
        Matrix res;
        memset(res.mat, , sizeof res.mat);
        for (int i = ; i < MAXN; i++)
            for (int j = ; j < MAXN; j++)
                for (int k = ; k < MAXN; k++)
                    res.mat[i][j] = (res.mat[i][j] + this->mat[i][k] * b.mat[k][j])%MOD;
        return res;
    }
}base, F0, FN;
Matrix fpow(Matrix base, ll n) {
    Matrix res;
    memset(res.mat, , sizeof res.mat);
    for (int i = ; i < MAXN; i++)
        res.mat[i][i] = ;
    while (n) {
        if (n&) res = res*base;
        base = base * base;
        n >>= ;
    }
    return res;
}
void init()
{
    base.mat[][] = ; base.mat[][] = ;
    base.mat[][] = ; base.mat[][] = ;
    memset(F0.mat, , sizeof F0.mat);
    F0.mat[][] = ; F0.mat[][] = ;
}
 
int main()
{
    int N;
    while (cin >> N) {
        if (N == -)
            break;
        init();
        FN = fpow(base, N);
        cout << FN.mat[][] << endl;
    }
    return ;
}