public class A {
public static void main(String[] args) {
int[] num = new int[5];
int max, min;
System.out.println("Numbers is:");
for (int i = 0; i < num.length; i++) {
num[i] = (int) (Math.random() * 900) + 100;
System.out.print(num[i] + " ");
}
max = min = num[0];
for (int i = 1; i < num.length; i++) {
if (max < num[i])
max = num[i];
if (min > num[i])
min = num[i];
}
System.out.println("\nMax: " + max);
System.out.println("Min: " + min);
}
}
2、編程輸出1-100間的整數,並且此整數必須滿足它是3的倍數,但不是5的倍數,也不是9的倍數,求這些數的和以及平均數。
public class B {
public static void main(String[] args) {
int sum = 0;
int count = 0;
for (int i = 1; i <= 100; i++) {
if ((i % 3 == 0) && (i % 5 != 0) && (i % 9 != 0)) {
3、5位數的密碼1XX23,其中百位和千位模糊不清,但知道該數能被57和67除盡,設計壹個算法,算出原有可能密碼。
public class C {
public static void main(String[] args) {
for (int i = 100; i <= 9900; i+=100) {
if (((10023 + i) % 57 == 0) || ((10023 + i) % 67 == 0)) {
System.out.println(i + 10023);
}
}
}
}
4、使用窮舉法輸出100以內所有素數。
public class D {
public static void main(String[] args) {
boolean bool;
for (int i = 2; i <= 100; i++) {
bool = true;
for (int j = 2; j <= (int)(Math.sqrt(i)); j++) {
if ((i % j) == 0) {
bool = false;
break;
}
}
if (bool == true) {
System.out.print(i + "\t");
}
}
}
}