The learning content:
1, judgment leap year
leap year rule: four years a leap, a hundred years does not run, then run four hundred years
given algorithm:
public class LeapYearQuestion {
static int isLeapYear(int year){
if((year%4==0)&&(year%100!=0)||year%400==0){
return 1;
}else{
return 0;
}
}
public static void main(String[] args){
System.out.print("输出2000~3000的所有闰年\n");
int count=0;
for(int i=2000;i<3000;i++){
if(isLeapYear(i)==1){
System.out.print(i+"\t");
count++;
if(count%10==0){
System.out.println("\n");
}
}
}
}
}
2. polynomial
2.1 one-dimensional polynomial evaluator
A polynomial of the general computing recursive algorithm methods can be used. First, the polynomial can be transformed into the following equivalent form:
P (X) = (... ((. 1 AN- X-AN 2 + X) * X +-AN. 3) + ... + A1 X) X + A0
can be seen from this expression, as long as one layer from the inside out in accordance with the following recursive manner, the entire value can be calculated to obtain a one-dimensional polynomial.
; = RN-AN. 1. 1-
Rk = Rk +. 1 + X AK;
K = n--2, ..., 1,0
latter layer a layer is calculated, the value of Ro is obtained by the polynomial P (x) of.
One-dimensional polynomial evaluation algorithm:
static double po1D(double a[],int n,double x){
double result = a[n-1];
for(int i=n-2;i>=0;i--){
result=result*x+a[i];
}
return result;
}
Example code:
public class Polynomial1D {
static double po1D(double a[],int n,double x){
double result = a[n-1];
for(int i=n-2;i>=0;i--){
result=result*x+a[i];
}
return result;
}
public static void main(String[] args) {
double[] a={-15,-7,7,2,-3,7,3};
double[] x={-2.0,-1,0,1,2,2.5};
for(int i=0;i<x.length;i++){
System.out.println("当x="+x[i]+"时,p(x)="+po1D(a,7,x[i]));
}
}
}
Random method 3.1 java language
3.1.1, Math.random () method, a random number is generated between 0 and 1 double, we can multiply it by a certain number, for example, multiplied by 100, which is less than the random number 100.
3.1.2, Random class java.util package itself, we can create a Random objects to generate random numbers, which can generate a random integer randomly float, randomized double, random long.
3.1.2, System class has a currentTimeMillis () method, which returns from at 0:00:00 on January 1, 1979 to present a number of milliseconds, the return type is long, we can take it as a random number.
Random generating random using the following method:
public class Random {
public static void main(String[] args){
java.util.Random r = new java.util.Random();
for(int i=0;i<10;i++){
for(int j=0;j<10;j++){
System.out.printf("%3d ",r.nextInt(100));
}
System.out.println();
}
}
}
3.2 Algorithm [0,1] uniformly distributed random numbers between the
First, setting a base base = 256.0, and two constants a = 17.0 and 189.0. Here, the base and generally an integral multiple of the base 2, the constants a and b can be freely taken empirically.
code show as below:
static double rand(double[] a){
double base,u,v,p,temp1,temp2,temp3;
base=256.0;
u=17.0;
v=189.0;
temp1 = (a[0])*u+v;
temp2 = (int)(temp1/base);
temp3 = temp1-temp2*base;
a[0]=temp3;
p=a[0]/base;
return p;
}
3.3 generating a random number range in any
Like the above-described method, but the required output bit configuration, the following is a detailed Code:
[m, n-] floating point random number between
public class Rand01 {
static double rand(double[] a){
double base,u,v,p,temp1,temp2,temp3;
base=256.0;
u=17.0;
v=189.0;
temp1 = (a[0])*u+v;
temp2 = (int)(temp1/base);
temp3 = temp1-temp2*base;
a[0]=temp3;
p=a[0]/base;
return p;
}
public static void main(String[] args){
double[] a= {5.0};
int m=50.0;
int n=100.0;
System.out.println("产生0~1的随机数如下：");
for(int i=0;i<7;i++){
System.out.printf("%d ",m+(n-m)*rand(a));
}
}
}
3.4 Algorithm uniformly distributed random integer between [m, n] of
With the algorithm uniformly distributed random number between [0, 1] above, if you want to get an integral, very simple, as long as the result to rounding.
Example code:
public class Rand01 {
static double rand(double[] a){
double base,u,v,p,temp1,temp2,temp3;
base=256.0;
u=17.0;
v=189.0;
temp1 = (a[0])*u+v;
temp2 = (int)(temp1/base);
temp3 = temp1-temp2*base;
a[0]=temp3;
p=a[0]/base;
return p;
}
public static void main(String[] args){
double[] a= {5.0};
int m=50;
int n=100;
System.out.println("产生0~1的随机数如下：");
for(int i=0;i<7;i++){
System.out.printf("%d ",m+(int)((n-m)*rand(a)));
}
}
}
4, complex operation
Complex arithmetic operations:
public class Plural {
//加法
static void cPlus(double a,double b,double c,double d, double[] e, double[] f){
e[0]=a+c;
f[0]=b+d;
}
//减法
static void cJian(double a, double b, double c, double d, double[] e, double[] f){
e[0]=a-c;
f[0]=b-d;
}
//乘法
static void cCheng(double a, double b, double c, double d, double[] e, double[] f){
e[0]=a*c-b*d;
f[0]=a*d+b*c;
}
//除法
static void cShu(double a, double b, double c, double d, double[] e, double[] f){
double sq;
sq = c*c+d*d;
e[0]=(a*c+b*d)/sq;
f[0]=(b*c-a*d)/sq;
}
static void cPowN(double a, double b, int n, double[] e, double[] f){
double result;
int i=0;
e[0]=a;
f[0]=b;
if(n==1){
return ;
}else{
for(i=1;i<n;i++){
cCheng(e[0],f[0],a,b,e,f);
}
}
}
public static void main(String[] args){
double a,b,c,d;
double[] e = {0.0};
double[] f = {0.0};
a=4;b=6;
c=2;d=-1;
cPlus(a,b,c,d,e,f);
System.out.println("( "+a+ "+"+b+"i"+") "+"+"+ "("+c+"+"+d+"i"+") ="+e[0]+"+"+f[0]+"i");
cJian(a,b,c,d,e,f);
System.out.println("( "+a+ "+"+b+"i"+") "+"-"+ "("+c+"+"+d+"i"+") ="+e[0]+"+"+f[0]+"i");
cCheng(a,b,c,d,e,f);
System.out.println("( "+a+ "+"+b+"i"+") "+"*"+ "("+c+"+"+d+"i"+") ="+e[0]+"+"+f[0]+"i");
cShu(a,b,c,d,e,f);
System.out.println("( "+a+ "+"+b+"i"+") "+"/"+ "("+c+"+"+d+"i"+") ="+e[0]+"+"+f[0]+"i");
}
}
5, factorial
5.1 cycle factorial
long fact(int n)
{
int i;
long result=1;
for(i=1;i<=n;i++)
{
result*=i;
}
return result;
}
5.2 uses recursion to calculate the factorial
long fact(int n){
if(n<=1)
return 1;
else
return n*fact(n-1);
}