用JAVA 编写程序,从键盘读入10个整数存入数组,输出最大值,最小值及它们所在数组中的位置 java如何编程实现从键盘输入10个数存入数组，显示出来并从...
java\u7f16\u7a0b\uff0c\u4ece\u952e\u76d8\u8f93\u516510\u6574\u6570\u5b58\u5165\u6570\u7ec4\u4e2d\uff0c\u8f93\u51fa\u8fd910\u4e2a\u6574\u6570\uff0c\u5e76\u6c42\u6700\u5c0f\u503c.import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int a = 10;
int[] b = new int[a];
int i=0;
int c = 0;
while(i<a){
b[i]=sc.nextInt();
if(i!=0){
c=(c>b[i])?b[i]:c;
}else{
c=b[i];
}
i++;
}
for(int s:b){
System.out.print(s+" ");
}
System.out.println();
System.out.println("\u6700\u5c0f\u6570\uff1a"+c);
}
}
import java.util.Scanner;
public class NumTest {
public static void main(String[] args) {
int[] nums=new int[10];
Scanner jin = new Scanner(System.in);
for (int i = 0; i < nums.length; i++) {
nums[i]=jin.nextInt();
}
int smallerNum=nums[0];
int biggerNum=nums[0];
System.out.print("\u8fd9\u5341\u4e2a\u6570\u662f\uff1a");
for (int i = 0; i < nums.length; i++) {
System.out.print(" "+nums[i]);
if (smallerNum>nums[i]) {
smallerNum=nums[i];
}
if (biggerNum<nums[i]) {
biggerNum=nums[i];
}
}
System.out.println();
System.out.println("\u6700\u5c0f\u503c\u4e3a\uff1a"+smallerNum);
System.out.println("\u6700\u5927\u503c\u4e3a\uff1a"+biggerNum);
}
}
import java.lang.*;
import java.util.StringTokenizer;
class array
{
public static void main(String args[])
{
StringTokenizer st=new StringTokenizer(\"10 20 30 40\");
String s=\"\";
int[] b=new int[10];
int Max=0,Min=0;
int count=0;
System.out.println(\"输入10个数字用空格隔开如:1 10 31 45 57 6 70 18 29 50\");
try
{
BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
s=in.readLine();
st=new StringTokenizer(s);
}
catch(Exception e)
{
System.out.print(e.getMessage());
}
while(st.hasMoreTokens())
{
b[count]=Integer.parseInt(st.nextToken());
count++;
}
Min=b[0];
for(int i=0;i<b.length;i++)
{
if(Max<b[i])
{
Max=b[i];
}
else if(Min>b[i])
{
Min=b[i];
}
}
System.out.println(\"Max:\"+Max);
System.out.println(\"Min:\"+Min);
}
}
这不难吧,一个循环就够了,自己试试.
这样的问题也要找高手？？？
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