从键盘上输入6个随机数存储在数组中,找出数组中最大值的下标位置。
package lianxi
;
import java
.util
.Scanner
;
public class Zzd {
public static void main(String
[] args
) {
int a
[]=new int[6];
int max
,i
;
try (Scanner sc
= new Scanner(System
.in
)) {
System
.out
.println("请输入6个值:");
for(i
=0;i
<a
.length
;i
++)
{
a
[i
]=sc
.nextInt();
}
}
max
=0;
for(i
=0;i
<a
.length
;i
++) {
if(a
[max
]<a
[i
])
{
max
=i
;
}
}
System
.out
.print("最大值为"+a
[max
]+"\n下标位置为"+max
);
}
}
有3个学生的成绩存储在数组中{78,89,73},计算并输出学生的平均成绩。
package lianxi
;
public class Pj {
public static void main(String
[] args
)
{
int a
[]=new int[] {78,89,73};
int sum
=0,pjz
;
for(int i
=0;i
<a
.length
;i
++) {
sum
=sum
+a
[i
];
}
pjz
=sum
/a
.length
;
System
.out
.print(pjz
);
}
}
将矩阵元素互换,存到另一个矩阵中,并输出
package lianxi
;
public class Jzzh {
public static void main(String
[] args
) {
int i
, j
;
int[][] a
={{1,2,3},{4,5,6}};
int[][] b
=new int[3][2];
for (i
=0;i
<a
.length
;i
++)
{
for (j
=0;j
<a
[i
].length
;j
++)
b
[j
][i
]=a
[i
][j
];
}
for (i
=0;i
<b
.length
;i
++) {
for (j
=0;j
<b
[i
].length
;j
++)
{
System
.out
.print(b
[i
][j
]+" ");
}
System
.out
.print("\n");
}
}
}
给定一个字符串,判断该字符串中是否包含某个子串.如果包含,求出子串的所有出现位置。 如:"abcd23abc34bcd"中,"bc"子串的出现位置为: 1,7,11.字符串和子串均由用户输入。
package lianxi
;
import java
.util
.Scanner
;
public class Zfc {
public static void main(String
[] args
) {
String s1
,s2
;
try (Scanner sc
= new Scanner(System
.in
)) {
System
.out
.println("请输入字符串:");
s1
=sc
.nextLine();
System
.out
.println("请输入要查询的子串:");
s2
=sc
.nextLine();
}
int a
=0,i
;
System
.out
.println("子串出现的位置为:");
while (true) {
i
=s1
.indexOf(s2
,a
);
if (i
==-1)
{
break;
}
System
.out
.print(i
+" ");
a
=i
+s2
.length();
}
}
}
输入一个由数字组成的字符串,统计出每个数字出现的次数。
package lianxi
;
import java
.util
.Scanner
;
public class Cxcs {
public static void main(String
[] args
) {
String s
;
int[] a
=new int[10];
try (Scanner sc
= new Scanner(System
.in
)) {
System
.out
.println("请输入字符串:");
s
=sc
.nextLine();
}
for(int i
=0;i
<10;i
++) {
String s1
=s
;
int index
=0;
while(true) {
index
=s1
.indexOf(String
.valueOf(i
));
if(index
!= -1){
s1
=s1
.substring(index
+1);
a
[i
]++;
}
else break;
}
System
.out
.print(i
+" 出现了"+a
[i
]+"次 ");
}
}
}
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