「计算njava」计算机二级考试时间
本篇文章给大家谈谈计算njava，以及计算机二级考试时间对应的知识点，希望对各位有所帮助，不要忘了收藏本站喔。
本文目录一览：
- 1、用java编写程序求n!
- 2、java 求N
- 3、用JAVA计算n!
- 4、用java语言计算N的数字根
- 5、java计算n的阶乘
用java编写程序求n!
import java.util.Scanner;
public class Text{
public static void main(String args[]){
System.out.println("请输入n");
Scanner sc = new Scanner(System.in);
int n= sc.nextInt();
int t=1;
int i=1;
while(i=n){
t=i*t;
i=i+1;
}
System.out.println("n!="+t);
}
}
//补充，JAVA文件保存时文件名必须和类名相同。
java 求N
对于
for(i=1;i=n;i++)){
fact=fact*i;
}
当i==1，fact = fact * 1 == 1 * 1
当i==2，fact = fact * 2 == 1 * 1 * 2
当i==3，fact = fact * 3 == 1 * 1 * 2 * 3
因为N!很大，所以用long表示还是嫌小。通常不会用int，long来表示。
会转义成char[]（人工模拟）或者如下。
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String args[]) {
Scanner scanner = new Scanner(System.in);
int count = 0;
BigInteger sum = null;
while (scanner.hasNext()) {
count = scanner.nextInt();
sum = BigInteger.valueOf(1);
for (int i = 1; i = count; i++) {
sum = sum.multiply(BigInteger.valueOf(i));
}
System.out.println(sum);
}
scanner.close();
}
}
如果需要重复的计算N！。
当N非常大时，每一次计算都会耗费非常多的时间。
所以会打一个表。用来保存计算过的N！。以空间换取时间。
确保在打表范围内，每一个N只会计算一次N！
package zhidao;
import java.math.BigInteger;
import java.util.Scanner;
public class N {
private static final int MAX = 1000;
private static BigInteger bis[] = new BigInteger[MAX];
public static void main(String args[]) {
Scanner scanner = new Scanner(System.in);
int n = 0;
BigInteger sum = null;
while (scanner.hasNext()) {
n = scanner.nextInt();
sum = BigInteger.valueOf(1);
// 方法1
// for (int i = 1; i = n; i++) {
// sum = sum.multiply(BigInteger.valueOf(i));
// }
sum = fact(n);
System.out.println(sum);
}
scanner.close();
}
public static BigInteger fact(int n) {
if (n == 0) {
bis[0] = BigInteger.valueOf(1);
return bis[0];
}
if (n MAX bis[n] != null) {
return bis[n];
}
System.out.println("calc" + n + "!");
BigInteger sum = BigInteger.valueOf(n).multiply(fact(n - 1));
bis[n] = sum;
return bis[n];
}
}
console：
20
calc20!
calc19!
calc18!
calc17!
calc16!
calc15!
calc14!
calc13!
calc12!
calc11!
calc10!
calc9!
calc8!
calc7!
calc6!
calc5!
calc4!
calc3!
calc2!
calc1!
2432902008176640000
20
2432902008176640000
22
calc22!
calc21!
1124000727777607680000
可以看到，计算过的数据就不会重复计算了。
但是这又引入了一个新的问题。
就是当递归的长度过长的时候会导致stackoverflow，这又是另一说了。
我来修正一个错误；
if (n MAX) {
bis[n] = sum;
}
return sum;
当n=1000时。
402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
用JAVA计算n!
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
/*
*作者：feihaiwuxian
*时间：May 16, 2009 11:10:49 AM
*作用：
*/
public class Calculate_N {
/**
* @param args
*/
private int result=1;
public void work_n(){
/********键盘输入数字并判断*********/
long n=1;
String flag;
BufferedReader br= null;
InputStreamReader is=new InputStreamReader (System.in);
br=new BufferedReader(is);
System.out.println("请输入数字(输入 0 结束)：\n");
/********键盘输入数字并判断*********/
while(true){
try {
flag=br.readLine();
n=Integer.parseInt(flag);
if(n==0){
System.out.println("程序退出！！");
break;
}else{
result=1;
for(int i=1;i=n;i++){
result*=i;//计算n!=1*2*3*4*....*n
}
System.out.println("n!=:"+result);
}
} catch (NumberFormatException e) {
System.out.println("输入错误，请输入数字！！");
} catch (IOException e) {
e.printStackTrace();
}
}
}
public static void main(String[] args) {
new Calculate_N().work_n();
}
}
一下是可以输入任意精度整数的：
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
/*
*作者：feihaiwuxian
*时间：May 16, 2009 11:10:49 AM
*作用：
*/
public class Calculate_N {
/**
* @param args
*/
private BigInteger result;//任意精度的整数
public void work_n(){
/********键盘输入数字并判断*********/
long n=1;
String flag;
BufferedReader br= null;
InputStreamReader is=new InputStreamReader (System.in);
br=new BufferedReader(is);
System.out.println("请输入数字(输入 0 结束)：\n");
/********键盘输入数字并判断*********/
while(true){
try {
flag=br.readLine();
n=Integer.parseInt(flag);
if(n==0){
System.out.println("\n程序退出！！");
break;
}else{
result=new BigInteger("1") ;
for(int i=1;i=n;i++){
result=result.multiply(new BigInteger(""+i));//计算n!=1*2*3*4*....*n
}
System.out.println("n!=:"+result);
}
} catch (NumberFormatException e) {
System.out.println("\n输入错误，请输入数字！！");
} catch (IOException e) {
e.printStackTrace();
}
}
}
public static void main(String[] args) {
new Calculate_N().work_n();
}
}
用java语言计算N的数字根
用java8的stream处理起来，代码更简洁些
可以参考：
public static void main(String[] args) {
String str = "1782";
while(str.length() != 1)
{
str = String.valueOf(Stream.of(str.split("")).mapToInt(Integer::valueOf).sum());
}
System.out.println(str);
}
中间那块函数含义也很简单
1. 先把这个字符分割成字符串数组
2. 字符串数组转化为对应stream
3. 把stream转化为IntStream
4. 直接用IntStream的sum方法求出和
5. 再把和转化为string，赋值给最开始的那个str
6. 然后判断str的长度是否为1，为1了就停止
java计算n的阶乘
int n=8;
int p=n;
for(int i=n;i1;i--){
p*=i;
}
print("%d\n",p);
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