矩阵的幂 斐波那契数列

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*Bmat 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 ^ nmat 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;}