黑马程序员技术交流社区

标题: java基础编程题 [打印本页]

作者: 15891252910 时间: 2016-8-3 09:48
标题: java基础编程题

3道JAVA基础编程练习题

[size=10.5000pt]【程序1】
[size=10.5000pt]题目：古典问题：有一对兔子，从出生后第3个月起每个月都生一对兔子，小兔子长到第三个月后每个月又生一对兔子，假如兔子都不死，问每个月的兔子[size=10.5000pt]对[size=10.5000pt]数为多少？
[size=10.5000pt]程序分析：兔子的规律为数列1,1,2,3,5,8,13,21....

[size=10.5000pt]public class Prog1{
[size=10.5000pt] [size=10.5000pt]public static void main(String[] args){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]int n = 10;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]System.out.println("第"+n+"个月兔子总数为"+fun(n));
[size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt] [size=10.5000pt]private static int fun(int n){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]if(n==1 || n==2)
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] return 1;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]else
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] return fun(n-1)+fun(n-2);
[size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt]}

[size=10.5000pt]【程序2】
[size=10.5000pt]题目：判断101-200之间有多少个素数，并输出所有素数。
[size=10.5000pt]程序分析：判断素数的方法：用一个数分别去除2到sqrt(这个数)，如果能被整除，则表明此数不是素数，反之是素数。

[size=10.5000pt]public class Prog2{
[size=10.5000pt] [size=10.5000pt]public static void main(String[] args){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]int m = 1;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]int n = 1000;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]int count = 0;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]//统计素数个数
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]for(int i=m;i<n;i++){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]if(isPrime(i)){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]count++;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]System.out.print(i+" ");
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]if(count%10==0){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]System.out.println();
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]System.out.println();
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]System.out.println("在"+m+"和"+n+"之间共有"+count+"个素数");
[size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt] [size=10.5000pt]//判断素数
[size=10.5000pt] [size=10.5000pt]private static boolean isPrime(int n){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]boolean flag = true;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]if(n==1)
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] flag = false;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]else{
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]for(int i=2;i<=Math.sqrt(n);i++){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]if((n%i)==0 || n==1){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]flag = false;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]break;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt] else
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt] flag = true;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] }
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]return flag;
[size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt]}

[size=10.5000pt]【程序3】
[size=10.5000pt]题目：打印出所有的"水仙花数"，所谓"水仙花数"是指一个三位数，其各位数字立方和等于该数本身。例如：153是一个"水仙花数"，因为153=1的三次方＋5的三次方＋3的三次方。
[size=10.5000pt]程序分析：利用for循环控制100-999个数，每个数分解出个位，十位，百位。

[size=10.5000pt]public class Prog3{
[size=10.5000pt] [size=10.5000pt]public static void main(String[] args){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]for(int i=100;i<1000;i++){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]if(isLotus(i))
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt] System.out.print(i+" ");
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]System.out.println();
[size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt] [size=10.5000pt]//判断水仙花数
[size=10.5000pt] [size=10.5000pt]private static boolean isLotus(int lotus){
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]int m = 0;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]int n = lotus;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]int sum = 0;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]m = n/100;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]n -= m*100;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]sum = m*m*m;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]m = n/10;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]n -= m*10;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]sum += m*m*m + n*n*n;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]if(sum==lotus)
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]return true;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]else
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt] [size=10.5000pt]return false;
[size=10.5000pt] [size=10.5000pt] [size=10.5000pt]}
[size=10.5000pt]}

[size=10.5000pt]

作者: 张彦虎 时间: 2016-8-3 09:51
乱码了

作者: 酷我之情 时间: 2016-8-3 09:53
什么东西煞笔

作者: 15891252910 时间: 2016-8-3 09:56

张彦虎发表于 2016-8-3 09:51
乱码了

尴尬尴尬，{:2_40:}

作者: zhaoyar 时间: 2016-8-3 13:16
？？？？

作者: 哈韩浪子 时间: 2016-8-18 10:28
什么都看不了啊能不能改一下啊

欢迎光临黑马程序员技术交流社区 (http://bbs.itheima.com/)

黑马程序员IT技术论坛 X3.2