示例代码¶

double max = a[0];
for (int i = 1; i < a.length; i++) {
    if (a[i] > max) max = a[i];
}

int N = a.length;
double sum = 0.0;
for (int i = 0; i < N; i++)
    sum += a[i];
double average = sum / N;

int N = a.length;
double[] b = new double[N];
for (int i = 0; i < N; i++)
    b[i] = a[i];

int N = a.length;
for (int i = 0; i < N / 2; i++) {
    double temp = a[i];
    a[i] = a[N-1-i];
    a[N-1-i] = temp;
}

int N = a.length;
double[][] c = new double[N][N];
for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
        // 计算 行i 和 列j 的点乘
        for (int k = 0; k < N ;k++) 
            c[i][j] += a[i][k] * b[k][j];
    }
}

public static int abs(int x) 
{
    if (x < 0) return -x;
    else    return x;
}

public static double abs(double x) {
    if (x < 0.0) return -x;
    else    return x;
}

public static boolean isPrime(int N) {
    if (N < 2) return false;
    for (int i = 2; i * i <= N; i++) {
        if (N % i == 0) return false;
    }
    return true;
}

public static double sqrt(double c) {
    if (c < 0) return Double.NaN;
    double err = 1e-15;
    double t = c;
    while (Math.abs(t - c / t) > err * t) {
        t = (c / t + t) / 2.0;
    }
    return t;
}

public static double hypotenuse(double a, double b) {
    return Math.sqrt(a * a + b * b);
}

public static double H(int N) {
    double sum = 0.0;
    for (int i = 1; i <= N; i++) {
        sum += 1.0 / i;
    }
    return sum;
}