矩阵的幂 斐波那契数列

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题意：

斐波那契数列是由如下递推式定义的数列

F0 = 0

F1 = 1

Fn+2 = Fn+1 + Fn

求这个数列第n项的值对10000取余后的结果

输入：

n = 10

输出：

55

#include <cstdio>
#include <vector>
using namespace std;
//用二维vector来表示矩阵
typedef vector<int> vec;
typedef vector<vec> mat;
typedef long long ll;

const int M = 10000;

//计算A*B
mat mul(mat &A, mat &B)
{
    mat C(A.size(), vec(B[0].size()));
    for (int i = 0; i < A.size(); i++){
        for (int k = 0; k < B.size(); k++){
            for (int j = 0; j < B[0].size(); j++){
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;
            }
        }
    }
    return C;
}

//计算A ^ n
mat pow(mat A, ll n)
{
    mat B(A.size(), vec(A.size()));
    for (int i = 0; i < A.size(); i++){
        B[i][i] = 1;
    }
    while (n > 0){
        if (n & 1)
            B = mul(B, A);
        A = mul(A, A);
        n >>= 1;
    }
    return B;
}

//输入
ll n;
void solve()
{
    mat A(2, vec(2));
    A[0][0] = 1;
    A[0][1] = 1;
    A[1][0] = 1;
    A[1][1] = 0;
    A = pow(A, n);
    printf("%d\n", A[1][0]);
}

int main()
{
    scanf("%d", &n);
    solve();
    return 0;
}

矩阵的幂 斐波那契数列

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原文地址：http://blog.csdn.net/a2459956664/article/details/51330448