方法一. 递推 

class Solution {
public:
    int fib(int n) {
        int MOD = 1e9 + 7;
        if (n < 2) return n;
        int p = 0, q = 0, r = 1;
        for (int i = 2; i <= n; i++) {
            p = q;
            q = r;
            r = (p + q) % MOD;
        }
        return r;
    }
};

方法二：矩阵快速幂

class Solution {
public:
    const int MOD = 1e9 + 7;
    
    int fib(int n) {
        if (n < 2) return n;
        vector<vector<long>> q{{1, 1},{1, 0}};
        vector<vector<long>> res = pow(q, n - 1);   
        return res[0][0];
    }
    // 快速幂：利用二进制表示法，将高次幂转化成二进制位为1处对应的各低次幂的乘积。
    vector<vector<long>> pow(vector<vector<long>>& a, int n) {
        vector<vector<long>> ret{{1, 0}, {0, 1}};   // 单位阵
        while (n) {
            if (n & 1) {
                ret = mutiply(ret, a);
            }
            n >>= 1;       
            a = mutiply(a, a);  
        }
        return ret;
    }
    // 定义矩阵乘法
    vector<vector<long>> mutiply(vector<vector<long>> &a, vector<vector<long>> &b) {
        vector<vector<long>> c{{0, 0}, {0, 0}};
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                c[i][j] = (a[i][0] * b[0][j] + a[i][1] * b[1][j]) % MOD;
            }
        }
        return c;
    }
};