Contents
- 1 (统计正教和负数的个数然后计算这些数的平均值）
- 2 (将千克转换成磅）编写程序，显示下面的表格（注意：1 千克为2.2 磅）
- 3 (将英里转換成千米）编写程序，显示下面的表格（注意：1 英里为1.609 千米）。
- 4 (千克与磅之间的互换）编写一个程序，并排显示下列两个表格。
- 5 (财务应用程序：计算将来的学费）
- 6 (找出最高分）
- 7 (找出两个分教最高的学生）
- 8 (找出能被5 和6 整除的數）
- 9 (找出能被5 或6 整除，但不能被两者同时整除的数）
- 10 (求满足{ n }^{ 2 }>12 000 的n 的最小值）使用while 循环找出满足n 大于12 000 的最小整数n
- 11 (求满足{ n }^{ 3}<12 000 的n 的最大值）用while 循环找出满足n3 小于12 000 的最大整数n。
- 12 (显示ACSII 码字符表）
- 13 (找出一个整数的因子）
- 14 (显示金字塔）
- 15 (打印金字塔形的教字）
- 16 (财务应用程序：比较不同利率下的贷款）
- 17 ( 示例抵消错误）
- 18 (数列求和）编写程序，计算下面数列的和：
- 19 (计算π）
编写一个程序，随机产生一个0到100 之间且包含0 和100 的整数。程序提示用户连续输人一个数字，直到它和计算机随机产生的数字相匹配为止。对用户每次输入的数字，程序都要告诉用户该输入值是偏大了，还是偏小了，这样用户可以明智地进行下一轮的猜测。
下面是一个运行示例：
What is 5 + 9? 12 Wrong answer. Try again. What is 5 + 9? 34 Wrong answer. Try again. What is 5 + 9? 14 You got it!
import java.util.Scanner;
public class GuessNumber {
public static void main(String[] args) {
// Generate a random number to be guessed
int number = (int)(Math.random() * 101);
Scanner input = new Scanner(System.in);
System.out.println("Guess a magic number between 0 and 100");
int guess = -1;
while (guess != number) {
// Prompt the user to guess the number
System.out.print("\nEnter your guess: ");
guess = input.nextInt();
if (guess == number)
System.out.println("Yes, the number is " + number);
else if (guess > number)
System.out.println("Your guess is too high");
else
System.out.println("Your guess is too low");
} // End of loop
}
}
(统计正教和负数的个数然后计算这些数的平均值）
编写程序，读入未指定个数的整数，判断读人的正数有多少个，读入的负数有多少个，然后计算这些输人值的总和及其平均值（不对 0 计数)。当输入为0 时，表明程序结束。将平均值以浮点数显示。下面是一个运行示例：
Enter an integer, the input ends if it is 0: 1 2 -1 3 0 The number of positives is 3 The number of negatives is 1 The total is 5.0 The average is 1.25
import java.util.Scanner;
public class Exercise_05_01 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int positives = 0; // Count the number of positive numbers
int negatives = 0; // Count the number of negative numbers
int count = 0; // Count all numbers
double total = 0; // Accumulate a totol
// Promopt the user to enter an integer or 0 to exit
System.out.print("Enter an integer, the input ends if it is 0: ");
int number = input.nextInt();
if (number == 0) { // Test for sentinel value
System.out.println("No numbers are entered except 0");
System.exit(1);
}
while (number != 0) {// Test for sentinel value
if (number > 0) {
positives++; // Increase positives
} else {
negatives++; // Increase negatives
}
total += number; // Accumulate total
count++; // Increase the count
number = input.nextInt();
}
// Calculate the average
double average = total / count;
// Display results
System.out.println(
"The number of positive is " + positives +
"\nThe number of negatives is " + negatives +
"\nThe total is total " + total +
"\nThe average is " + average);
}
}
(将千克转换成磅）编写程序，显示下面的表格（注意：1 千克为2.2 磅）
/*
(Conversion from kilograms to pounds) Write a program that displays the following
table (note that 1 kilogram is 2.2 pounds):
Kilograms Pounds
1 2.2
3 6.6
...
197 433.4
199 437.8
*/
public class Exercise_05_02 {
public static void main(String[] args) {
final double POUNDS_PER_KILOGRAM = 2.2; // Create constant
// Display header for table
System.out.println("Kilograms Pounds");
// Display table
for (int i = 1; i <= 199; i += 2) {
System.out.printf(
"%-15d%6.1f\n", i, (i * POUNDS_PER_KILOGRAM));
}
}
}
(将英里转換成千米）编写程序，显示下面的表格（注意：1 英里为1.609 千米）。
/*
(Conversion from miles to kilometers) Write a program that displays the following
table (note that 1 mile is 1.609 kilometers):
Miles Kilometers
1 1.609
2 3.218
...
9 14.481
10 16.090
*/
public class Exercise_05_03 {
public static void main(String[] args) {
// Create a constant number of kilometers in a mile
final double KILOMETERS_PER_MILE = 1.609;
// Display table header
System.out.println(
"Miles Kilometers");
// Create and display table showing conversion from miles to kilometers
for (int i = 1; i <= 10; i++) {
System.out.printf(
"%-13d%-10.3f\n", i, (i * KILOMETERS_PER_MILE));
}
}
}
(千克与磅之间的互换）编写一个程序，并排显示下列两个表格。
public class Exercise_05_04 {
public static void main(String[] args) {
// Create constant value for number of pounds per kilogram
final double POUNDS_PER_KILOGRAM = 2.2;
// Display table header
System.out.println(
"Kilograms Pounds | Pounds Kilograms");
for (int k = 1, p = 20; k <= 199 && p <= 515; k += 2, p += 5) {
System.out.printf(
"%-12d%7.1f", k, (k * POUNDS_PER_KILOGRAM));
System.out.print(" | ");
System.out.printf(
"%-9d%12.2f\n",
p, (p / POUNDS_PER_KILOGRAM));
}
}
}
(财务应用程序：计算将来的学费）
假设今年某大学的学费为10 000 美元，学费的年增长率为5%。— 年后，学费将是10 500 美元。编写程序，计算丨0 年后的学费.以及从现在开始的丨0 年后算起，4 年内总学费是多少？
/*
(Financial application: compute future tuition) Suppose that the tuition for a university
is $10,000 this year and increases 5% every year. In one year, the tuition
will be $10,500. Write a program that computes the tuition in ten years and the
total cost of four years’ worth of tuition after the tenth year.
*/
public class Exercise_05_07 {
public static void main(String[] args) {
int totalCost = 0; // Accumulate total cost of four years tution
int tuition = 10000;
int tuitionTenthYear;
for (int year = 1; year <= 14; year++) {
// Increase tution by 5% every year
tuition += (tuition * 0.05);
if (year > 10) // Test for after 10 years
totalCost += tuition; // Accumulate tution cost
// Cost of tution in ten years
if (year == 10)
tuitionTenthYear = tuition;
}
// Display the cost of tution in ten years
System.out.println("Tuition in ten years: $" + tuitionTenthYear);
// Display the cost of four years' worth of tution after tenth year
System.out.println("Total cost for four years' worth of tuition" +
" after the tenth year: $" + totalCost);
}
}
(找出最高分）
编写程序，提示用户输人学生的个数、每个学生的名字及其分数，最后显示得最高分的学生的名字。
import java.util.Scanner;
public class Exercise_05_05 {
public static void main(String[] args) {
// Create Scanner object
Scanner input = new Scanner(System.in);
int highestScore = 0; // Holds the hightest student score
String highestScoreName = ""; // Holds the student name with highest score
// Prompt the user to enter the number of students
System.out.print("Enter the number of students: ");
int numberOfStudents = input.nextInt();
// Prompt the user to enter each student's name and score
System.out.println("Enter each student’s name and score");
for (int i = 0; i < numberOfStudents; i++) {
System.out.print(
"Student: " + (i + 1) +
"\n Name: ");
String name = input.next();
System.out.print(
" Score: ");
int score = input.nextInt();
// Test if score is higher than highest score
if (score > highestScore) {
highestScore = score;
highestScoreName = name;
}
}
// Display the name of the student with the highest score
System.out.println("Student with the highest score: " + highestScoreName);
}
}
(找出两个分教最高的学生）
编写程序，提示用户输人学生的个数、每个学生的名宇及其分数，最后显示获得最高分的学生和第二高分的学生。
import java.util.Scanner;
public class Exercise_05_06 {
public static void main(String[] args) {
// Create a Scanner
Scanner input = new Scanner(System.in);
// Prompt the user to enter the number of students
System.out.print("Enter the number of students: ");
int numberOfStudents = input.nextInt();
int score, // Holds students' score
highest = 0, // Highest score
secondHigest = 0; // Second highest score
String name = "", // Holds students' name
student1 = "", // Highest scoring student name
student2 = ""; // Second highest scoring student name
// Prompt the user to enter each students' name and score
System.out.println("Enter each students' name and score:");
for (int i = 0; i < numberOfStudents; i++) {
System.out.print(
"Student: " + (i + 1) + "\n Name: ");
name = input.next();
System.out.print(" Score: ");
score = input.nextInt();
if (i == 0) {
// Make the first student the highest scoring student so far
highest = score;
student1 = name;
} else if (i == 1 && score > highest) {
// Second student entered scored
// higher than first student
secondHigest = highest;
highest = score;
student2 = student1;
student1 = name;
} else if (i == 1) {
// Second student entered scored
// lower than first student
secondHigest = score;
student2 = name;
} else if (i > 1 && score > highest && score > secondHigest) {
// Last student entered has the highest score
secondHigest = highest;
student2 = student1;
highest = score;
student1 = name;
} else if (i > 1 && score > secondHigest) {
// Last student entered has the second highest score
student2 = name;
secondHigest = score;
}
}
// Display the student with the hightest score
// and the student with the second-hightest score.
System.out.println(
"Higest scoring student: " + student1 +
"\nSecond Higest scoring student: " + student2);
}
}
(找出能被5 和6 整除的數）
编写程序，显示从100 到1000 之间所有能被5 和6 整除的数，每行显示10 个。数字之间用一个空格字符隔开。
public class Exercise_05_07 {
public static void main(String[] args) {
final int NUMBERS_PER_LINE = 10; // Display 10 numbers per line
int count = 0; // Count the number of numbers divisible by 5 and 6
// Test all numbers from 100 to 1,000
for (int i = 100; i <= 1000; i++) {
// Test if number is divisible by 5 and 6
if (i % 5 == 0 && i % 6 == 0) {
count++; // increment count
// Test if numbers per line is 10
if (count % NUMBERS_PER_LINE == 0) {
System.out.println(i);
} else {
System.out.print(i + " ");
}
}
}
}
}
(找出能被5 或6 整除，但不能被两者同时整除的数）
编写程序，显示从100 到200 之间所有能被5 或6 整除，但不能被两者同时整除的数，每行显示 10 个数。数字之间用一个空格字符隔开
public class Exercise_05_08 {
public static void main(String[] args) {
final int NUMBERS_PER_LINE = 10; // Number of numbers in each line displayed
int count = 0; // Count the number divisible by 5 or 6, but not both
for (int i = 100; i <= 200; i++) {
if (i % 5 == 0 ^ i % 6 == 0) {
count++;
if (count % NUMBERS_PER_LINE == 0) {
System.out.println(i);
} else {
System.out.print(i + " ");
}
}
}
System.out.println();
}
}
(求满足{ n }^{ 2 }>12 000 的n 的最小值）使用while 循环找出满足n 大于12 000 的最小整数n
public class Exercise_05_09 {
public static void main(String[] args) {
int n = 0; // Start n at 0
// Find the smallest n such that n^2 > 12,000
while (Math.pow(n, 2) < 12000) {
n++; // Increment n
}
// Display result
System.out.println(
"The smallest integer n such that n^2 is greater than 12,000: " + n);
}
}
(求满足{ n }^{ 3}<12 000 的n 的最大值）用while 循环找出满足n3 小于12 000 的最大整数n。
public class Exercise_05_10 {
public static void main(String[] args) {
int n = 0; // Intialize n at 0;
// Find the largest n such that n^3 is < 12,000
while (Math.pow(n + 1, 3) < 12000) {
n++; // Increment n
}
// Display result
System.out.println(
"Largest integer n such that n^3 is less than 12,000: " + n);
}
}
(显示ACSII 码字符表）
编写一个程序，打印ASCII 字符表从到的字符。每行打印10 个字符。ASCII 码表如图所示。数字之间用一个空格字符隔开。
public class Exercise_05_11 {
public static void main(String[] args) {
// Number of characters per line
final int NUMBER_OF_CHARACTERS_PER_LINE = 10;
int count = 0; // Count the number of characters
// Print the ASCII characters from ! to ~
for (int i = 33; i <= 126; i++) {
count++; // Increment count
// Display 10 characters per line
if (count % 10 == 0) {
System.out.println((char) i);
} else {
System.out.print((char) i + " ");
}
}
System.out.println();
}
}
(找出一个整数的因子）
编写程序，读入一个整数，然后以升序显示它的所有最小因子。例如,若输人的整数是120, 那么输出就应该是：2, 2, 2, 3, 5。
import java.util.Scanner;
public class Exercise_05_12 {
public static void main(String[] args) {
// Create a Scanner
Scanner input = new Scanner(System.in);
// Prompt the user to enter an integer
System.out.print("Enter an integer: ");
int number = input.nextInt();
int index = 2; // Numbers to test as factors
// Find and display all the smallest factors in increasing order
while (number / index != 1) {
// Test as a factor of number
if (number % index == 0) {
System.out.print(index + ", ");
number /= index;
} else {
index++; // Increment index
}
}
System.out.println(number + ".");
}
}
(显示金字塔）
编写程序，提示用户输人一个在1到15 之间的整数，然后显示一个金字塔形状的图案，如下面的运行示例所示：
import java.util.Scanner;
public class Exercise_05_13 {
public static void main(String[] args) {
// Create a Scanner object
Scanner input = new Scanner(System.in);
// Prompt the user to enter an integer from 1 to 15
System.out.print("Enter the number of lines: ");
int numberOfLines = input.nextInt();
// Display pyramid
for (int rows = 1; rows <= numberOfLines; rows++) {
// Create spaces in each row
for (int s = numberOfLines - rows; s >= 1; s--) {
System.out.print(" ");
}
// Create decending numbers in each row
for (int l = rows; l >= 2; l--) {
System.out.print(l + " ");
}
// Create ascending number in each row
for (int r = 1; r <= rows; r++) {
System.out.print(r + " ");
}
// End line
System.out.println();
}
}
}
(打印金字塔形的教字）
编写一个嵌套的for 循环，打印下面的输出
public class Exercise_05_14 {
public static void main(String[] args) {
int startRight = 0, // Initialize decending numbers
endSpace = 7; // Initialize number of white space in row
// Display number of rows and numbers in each row
for (int row = 1; row <= 128; row += row) {
// Display white space
for (int startSpace = 0; startSpace < endSpace; startSpace++) {
System.out.print(" ");
}
// Display acending numbers
for (int l = 1; l <= row; l += l) {
System.out.printf("%4d", (l));
}
// Display decending numbers
for (int r = startRight; r > 0; r /= 2) {
System.out.printf("%4d", (r));
}
System.out.println(); // End line
endSpace--; // Decrement endSpace
startRight = row; // Assign row to startRight
}
}
}
(财务应用程序：比较不同利率下的贷款）
编写程序，让用户输入贷款总额和以年为单位的贷款期限，然后显示利率从5% 到8%, 每次递增1/8 的过程中，每月的支付额和总支付额。下面是一个运行示例
import java.util.Scanner;
public class Exercise_05_15 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
// Prompt the user to enter the loan amount and
// loan period in number of years
System.out.print("Loan Amount: ");
double loanAmount = input.nextDouble();
System.out.print("Number of Years: ");
int numberOfYears = input.nextInt();
// Display table header
System.out.println(
"Interest Rate Monthly Payment Total Payment");
// Display table with interest rates
for (double i = 5.0; i <= 8; i += 0.125) {
System.out.printf("%-5.3f", i);
System.out.print("% ");
double monthlyInterestRate = i / 1200;
double monthlyPayment = loanAmount * monthlyInterestRate / (1
- 1 / Math.pow(1 + monthlyInterestRate, numberOfYears * 12));
System.out.printf("%-19.2f", monthlyPayment);
System.out.printf("%-8.2f\n", (monthlyPayment * 12) * numberOfYears);
}
}
}
( 示例抵消错误）
当处理一个很大的数字以及一个很小的数字的时候，会产生一个抵消错误( cancellation error)。例如，100 000 000.0 + 0.000 000 001 等于100 000 000.0。为了避免抵消错误，从而获得更加精确的结果，谨慎选择计算的次序。比如，在计算下面的数列时，从右到左计算要比从左到右计算得到的结果更精确：
1+\cfrac { 1 }{ 2 } +\frac { 1 }{ 3 } +…+\frac { 1 }{ n }编写程序对上面的数列从左到右和从右到左计算的结果进行比较，这里取n=50000。
public class Exercise_05_16 {
public static void main(String[] args) {
// Compute series from left to right
double sumLeftToRight = 0.0;
for (double i = 1.0; i <= 50000.0; i++) {
sumLeftToRight += 1 / i;
}
// Display result of series sum from left to right
System.out.println("Summation of series left to right: " + sumLeftToRight);
// Compute series from right to left
double sumRightToLeft = 0.0;
for (double i = 50000.0; i >= 1.0; i--) {
sumRightToLeft += 1 / i;
}
// Display result of series sum from right to left
System.out.println("Summation of series right to left: " + sumRightToLeft);
// Compare the results
System.out.println("Summation of the series right to left - "
+ "Summation of the series left to right: "
+ (sumRightToLeft - sumLeftToRight));
}
}
(数列求和）编写程序，计算下面数列的和：
1+\cfrac { 3 }{ 5 } +\frac { 5 }{ 7 } +\frac { 7 }{ 9 } +\frac { 9 }{ 11 } …+\frac { 97 }{ 99 }public class Exercise_05_17 {
public static void main(String[] args) {
// Sum the series
double sum = 0.0;
for (double n = 1.0; n <= 97.0; n += 2) {
sum += n / (n + 2);
}
System.out.println(
"Series: 1 / 3 + 3 / 5 + 5 / 7 + 7 / 9 + 9 / 11 + 11 / 13 + " +
" ... + 95 / 97 + 97 / 99");
System.out.println("Sum of series: " + sum);
}
}
(计算π）
使用下面的数列可以近似计算π:
编写程序. 显示当i=10000. 20000 100000 时 π 的值
public class Exercise_05_18 {
public static void main(String[] args) {
// Compute PI value for i = 10000,
double sum = 0;
double value = 10000.0;
for (double d = 1; d <= (2 * value - 1); d += 2) {
sum += 1 / d;
d += 2;
sum -= 1 / d;
}
double pi = 4 * sum;
// Display result
System.out.println("PI value for i = 10000: " + pi);
// Compute PI value for i = 20000,
sum = 0;
value = 20000.0;
for (double d = 1; d <= (2 * value - 1); d += 2) {
sum += 1 / d;
d += 2;
sum -= 1 / d;
}
pi = 4 * sum;
// Display result
System.out.println("PI value for i = 20000: " + pi);
// Compute PI value for i = 20000,
sum = 0;
value = 100000.0;
for (double d = 1; d <= (2 * value - 1); d += 2) {
sum += 1 / d;
d += 2;
sum -= 1 / d;
}
pi = 4 * sum;
// Display result
System.out.println("PI value for i = 100000: " + pi);
}
}