第一题
示例代码
#include <stdio.h>
#include <stdlib.h> // 用于使用exit()函数int main() {double maxNumber = -1; // 初始设为-1，以便于第一个非负数可以直接成为当前最大值double currentNumber;// 循环读取用户输入while (1) {printf("Enter a number: ");if (scanf_s("%lf", ¤tNumber) != 1) { // 检查输入是否正确printf("输入错误，请输入有效的数字。\n");fflush(stdin); // 清除错误输入continue;}// 如果输入的是0或负数，则退出循环if (currentNumber <= 0) {break;}// 如果当前输入的数大于已知的最大数，则更新最大数if (currentNumber > maxNumber) {maxNumber = currentNumber;}}// 输出结果if (maxNumber == -1) {printf("没有输入任何非负数。\n");}else {printf("The largest number entered was %lf\n", maxNumber);}return 0;
}输出
第二题
示例代码
#include <stdio.h>// 定义一个函数来计算两个数的最大公约数
int gcd(int a, int b) {while (b != 0) {int temp = b;b = a % b;a = temp;}return a;
}int main() {int num1, num2;printf("Enter two intergers: ");if (scanf_s("%d %d", &num1, &num2) != 2) {printf("输入错误，请输入有效的整数。\n");return 1; // 非正常退出}// 调用gcd函数计算最大公约数int result = gcd(num1, num2);// 显示结果printf("Greatest common divisor: %d\n", result);return 0;
}输出
第三题
示例代码
#include <stdio.h>// 定义一个函数来计算两个数的最大公约数
int gcd(int a, int b) {while (b != 0) {int temp = b;b = a % b;a = temp;}return a;
}int main() {int numerator, denominator;printf("Enter a fraction: ");if (scanf_s("%d/%d", &numerator, &denominator) != 2) {printf("输入错误，请输入有效的分数形式（分子/分母）。\n");return 1; // 非正常退出}// 计算最大公约数int commonDivisor = gcd(numerator, denominator);// 化简分数int simplifiedNumerator = numerator / commonDivisor;int simplifiedDenominator = denominator / commonDivisor;// 显示结果if (simplifiedDenominator == 1) {printf("In lowest terms: %d\n", simplifiedNumerator);}else {printf("In lowest terms: %d/%d\n", simplifiedNumerator, simplifiedDenominator);}return 0;
}输出
第四题
5.2节broker.c代码如下
#include<stdio.h>int main()
{float commission, value;int num;printf("Enter value of trade: ");scanf_s("%f", &value);if (value < 2500.00f){commission = 30.00f + .017f * value;}else if (value < 6250.00f){commission = 56.00f + .0066f * value;}else if (value < 20000.00f){commission = 76.00f + .0034f * value;}else if (value < 50000.00f){commission = 100.00f + .0022f * value;}else if (value < 500000.00f){commission = 155.00f + .0011f * value;}else{commission = 255.00f + .0009f * value;}if (commission < 39.00f){commission = 39.00f;}printf("Commission: $%.2f\n", commission);return 0;
}题目分析：这个结合本篇第一题的解法，加入while循环，循环条件为真，当判断输入为0时跳出循环即可，不难给出代码。
示例代码
#include<stdio.h>int main()
{float commission, value;while (1){printf("Enter value of trade: ");if (scanf_s("%f", &value) != 1) {printf("输入错误，请输入有效的交易金额。\n");continue; // 继续下一次循环}if (value!=0){if (value < 2500.00f){commission = 30.00f + .017f * value;}else if (value < 6250.00f){commission = 56.00f + .0066f * value;}else if (value < 20000.00f){commission = 76.00f + .0034f * value;}else if (value < 50000.00f){commission = 100.00f + .0022f * value;}else if (value < 500000.00f){commission = 155.00f + .0011f * value;}else{commission = 255.00f + .0009f * value;}if (commission < 39.00f){commission = 39.00f;}printf("Commission: $%.2f\n\n", commission);}else break;}return 0;
}输出
第五题
示例代码
#include <stdio.h>int main() {long number;long reversed = 0;int digit;printf("请输入一个整数：");if (scanf_s("%ld", &number) != 1) {printf("输入错误，请输入有效的整数。\n");return 1; // 非正常退出}// 处理负数long originalNumber = number;if (number < 0) {number = -number;}do {digit = number % 10; // 获取当前的最低位reversed = reversed * 10 + digit; // 将数字添加到结果中number /= 10; // 去掉最低位} while (number != 0);// 如果原始输入是负数，反转的结果也应该带上负号if (originalNumber < 0) {reversed = -reversed;}printf("逆序后的数字是：%ld\n", reversed);return 0;
}输出
第六题
示例代码
#include <stdio.h>int main() {int n;int i;printf("Enter a integer：");if (scanf_s("%d", &n) != 1) {printf("输入错误，请输入有效的整数。\n");return 1; // 非正常退出}// 检查 n 是否大于等于 4，因为 2 是最小的偶数，2^2=4if (n < 4) {printf("n 至少应该是 4 或更大。\n");return 1; // 非正常退出}// 使用 for 循环从 2 到 n 步长为 2 （只取偶数）for (i = 2; i * i <= n; i += 2) {// 计算平方并输出int square = i * i;printf("%d 的平方是 %d\n", i, square);}return 0;
}输出
