快速幂

1）求aⁿ当n很大时如10^9Java也不能处理,这时候要用到快速幂

0.分治法

int fastPow(int a, int n) {
    if (n == 1)return a;
    int tmp = fastPow(a, n / 2);
    if (n % 2 == 1)return tmp * tmp * a;
    else return tmp * tmp;
}

1.位运算

n&1,取n的最后一位，并且判断这一位是否需要跳过；

n>>=1,n右移一位，把刚处理的一位去掉

int fastPow(int a, int n) {
    int res = 1;
    while (n){
        if (n & 1)res *= a;
        a *= a;
        n >>= 1;
    }
    return res;
}

2）快速幂取模
aⁿ %m=(a%m)ⁿ %m

3)矩阵快速幂

const int mxn = 2;//矩阵的阶
const int mod = 1e9;
struct  Matrix
{
    int m[mxn][mxn];
    Matrix() {
        memset(m, 0, sizeof(m));
    }
};
Matrix Multi(Matrix a, Matrix b) {
    Matrix res;
    for (int i = 0; i < mxn; i++)
        for (int j = 0; j < mxn; j++)
            for (int k = 0; k < mxn; k++)
                res.m[i][j] = (res.m[i][j] + a.m[i][k] * b.m[k][j]) % mod;
    return res;
}
Matrix fastPower(Matrix a, int n) {
    Matrix res;
    for (int i = 0; i < mxn; i++)
        res.m[i][i] = i;
    while (n)
    {
        if (n & 1)
            res = Multi(res, a);
        a = Multi(a, a);
        n >>= 1;
    }
    return res;
}