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C语言程序设计：现代设计方法习题笔记《chapter3》

article 2024/10/24 7:47:19

第一题

代码示例：

#include<stdio.h>

int main()
{
	printf("Enter a date（mm/dd/yyyy）: ");
	int day, month, year;
	scanf_s("%d/%d/%d", &month, &day, &year);
	printf("%04d%02d%02d", year, month, day);
	return 0;
}

输出：

如果输出格式这样设置：

printf("%4d%2d%2d", year, month, day);

输出将为：

可见"d"前面的数字表示所占的位数，如果小于指定的位数，在前面加数字0，则会在高位补0，如果不加数字0，则为空，保留占位符。

第二题

示例代码：

#include<stdio.h>

int main()
{
	printf("Enter item number: ");
	int item;
	scanf_s("%d", &item);

	printf("Enter uint price: ");
	float price;
	scanf_s("%f", &price);

	printf("Enter purchase date: ");
	int month, day, year;
	scanf_s("%d/%d/%d", &month, &day, &year);

	printf("Item\t\tUint\t\tPurchase\n");
	printf("\t\tPrice\t\tDate\n");
	printf("%d\t\t$  %.2f\t%02d/%02d/%04d", item, price, month, day, year);
	return 0;
}

输出：

第三题

示例代码：

#include<stdio.h>

int main()
{
	printf("Enter ISBN: ");
	int prefix, identifier, code, digit;
	long number;
	scanf_s("%d-%d-%d-%ld-%d", &prefix, &identifier, &code, &number, &digit);

	printf("GS1 prefix: %d\n", prefix);
	printf("Group identifier: %d\n", identifier);
	printf("Publisher code: %d\n", code);
	printf("Item number: %ld\n", number);
	printf("Check digit: %d\n", digit);
	return 0;
}

输出：

第四题

示例代码：

#include<stdio.h>
int main()
{
	printf("Enter phone number [(xxx) xxx-xxxx]: ");
	int a, b, c;
	scanf_s("(%d) %d-%d", &a, &b, &c);
	printf("You entered %d.%d.%d", a, b, c);
	return 0;
}

输出：

第五题

示例代码：

#include<stdio.h>

int main()
{
	int a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p;
	printf("Enter the numbers from 1 to 16 in any order: \n");
	scanf_s("%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d", &a, &b, &c, &d, &e, &f, &g, &h, &i, &j, &k, &l, &m, &n, &o, &p);
	printf("%d\t%d\t%d\t%d\n", a, b, c, d);
	printf("%d\t%d\t%d\t%d\n", e, f, g, h);
	printf("%d\t%d\t%d\t%d\n", i, j, k, l);
	printf("%d\t%d\t%d\t%d\n", m, n, o, p);

//row sum
	int rowSum1, rowSum2, rowSum3, rowSum4;
	int columnSum1, columnSum2, columnSum3, columnSum4;
	int diagSum1, diagSum2;
	rowSum1 = a + b + c + d;
	rowSum2 = h + e + f + g;
	rowSum3 = k + i + j + l;
	rowSum4 = m + n + o + p;

	//column sum
	columnSum1 = a + e + i + m;
	columnSum2 = b + f + j + n;
	columnSum3 = c + g + k + o;
	columnSum4 = d + h + l + p;

	//diag sum
	diagSum1 = a + f + k + p;
	diagSum2 = d + j + g + m;

	printf("Row sums: %d %d %d %d\n", rowSum1, rowSum2, rowSum3, rowSum4);
	printf("Column sums: %d %d %d %d\n", columnSum1, columnSum2, columnSum3, columnSum4);
	printf("Diagonal sums: %d %d\n", diagSum1, diagSum2);

	return 0;
}

输出：

上面代码输出是左对齐的，而书中给出的示例是右对齐，虽然没有明确要求，但是这里给出实现代码：

#include<stdio.h>

int main()
{
	int a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p;
	printf("Enter the numbers from 1 to 16 in any order: \n");
	scanf_s("%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d%d", &a, &b, &c, &d, &e, &f, &g, &h, &i, &j, &k, &l, &m, &n, &o, &p);
	printf("%2d\t%2d\t%2d\t%2d\n", a, b, c, d);
	printf("%2d\t%2d\t%2d\t%2d\n", e, f, g, h);
	printf("%2d\t%2d\t%2d\t%2d\n", i, j, k, l);
	printf("%2d\t%2d\t%2d\t%2d\n", m, n, o, p);

//row sum
	int rowSum1, rowSum2, rowSum3, rowSum4;
	int columnSum1, columnSum2, columnSum3, columnSum4;
	int diagSum1, diagSum2;
	rowSum1 = a + b + c + d;
	rowSum2 = h + e + f + g;
	rowSum3 = k + i + j + l;
	rowSum4 = m + n + o + p;

	//column sum
	columnSum1 = a + e + i + m;
	columnSum2 = b + f + j + n;
	columnSum3 = c + g + k + o;
	columnSum4 = d + h + l + p;

	//diag sum
	diagSum1 = a + f + k + p;
	diagSum2 = d + j + g + m;

	printf("Row sums: %d %d %d %d\n", rowSum1, rowSum2, rowSum3, rowSum4);
	printf("Column sums: %d %d %d %d\n", columnSum1, columnSum2, columnSum3, columnSum4);
	printf("Diagonal sums: %d %d\n", diagSum1, diagSum2);

	return 0;
}

输出：

可以发现已经和书中一致。

第六题

示例代码：

#include<stdio.h>

int main()
{
	int num1, denom1, num2, denom2, result_num, result_denom;
	printf("Enter two fractions separated by a plus sign: ");
	scanf_s("%d/%d+%d/%d", &num1, &denom1, &num2, &denom2);
	result_denom = denom1 * denom2;
	result_num = num1 * denom2 + num2 * denom1;
	printf("\nThe sum is: %d/%d", result_num, result_denom);
	return 0;
}

输出：