【403】COMP9024 Exercise

Week 1 Exercises

fiveDigit.c

There is a 5-digit number that satisfies 4 * abcde = edcba, that is,when multiplied by 4 yields the same number read backwards.Write a C-program to find this number.

int swap_num(int a)
{
	int result = 0;
 
	while (1)
	{
		int i = a % 10;
		result = result * 10 + i;
		a = a / 10;
		if (a == 0)
			break;
	}
	return result;
}

参考：c语言编程:实现数字的翻转

innerProdFun.c

Write a C-function that returns the inner product of two n-dimensional vectorsa and b, encoded as 1-dimensional arrays of n floating point numbers.

Use the function prototype float innerProduct(float a[], float b[], int n).

By the way, the inner product of two vectors is calculated by the sum for i=1..n of a_i * b_i

float innerProduct(float a[], float b[], int n)
{
	int i;
	float sep, sum = 0.0;
	for (i = 0; i < n; i++)
	{
		sep = a[i] * b[i];
		sum = sum + sep;
	}
	return sum;
}

matrixProdFun.c

Write a C-function to compute C as the matrix product of matrices A and B.

Use the function prototype void matrixProduct(float a[M][N], float b[N][P], float c[M][P])

You can assume that M, N, P are given as symbolic constants, e.g.

#define M 3
#define N 4
#define P 4

By the way, the product of an m x n matrix A and an n x p matrix B is the m x p matrix C such that C_ij is the sum for k=1..n of A_ik * B_kj for all i∈{1..m} and j∈{1..p}

void matrixProduct(float a[M][N], float b[N][P], float c[M][P])
{
	float innerProduct(float a[], float b[], int n);
 
	int i, j, k;
	float sum;
	float rr[N], cc[N];
 
	for (i = 0; i < M; i++)
	{
		for (j = 0; j < N; j++)
			rr[j] = a[i][j];
 
		for (k = 0; k < P; k++)
		{
			for (j = 0; j < N; j++)
				cc[j] = b[j][k];
 
			sum = innerProduct(rr, cc, N);
			c[i][k] = sum;
		}
	}
}

数据调用

#include <stdio.h>
 
#define M 2
#define N 3
#define P 2
 
int main()
{
	void matrixProduct(float a[M][N], float b[N][P], float c[M][P]);
	float innerProduct(float a[], float b[], int n);
 
	float a[2][3] = { 1, 2, 3, 4, 5, 6 };
	float b[3][2] = { 1, 2, 3, 4, 5, 6 };
 
	float c[2][2];
	matrixProduct(a, b, c);
 
	int i, j;
	for (i = 0; i < 2; i++)
	{
		for (j = 0; j < 2; j++)
		{
			printf("%0.0f  ", c[i][j]);
		}
		printf("\n");
	}
 
	return 0;
}

able.c

Write a C-program that outputs, in alphabetical order, all strings that use each of the characters 'a', 'b', 'l', 'e' exactly once.

How many strings are there actually?

void able()
{
	char a = 'a';
	char b = 'b';
	char l = 'l';
	char e = 'e';
	char z = 'z';
 
	char rr[26];
	char ss[26];
 
	int i, j, tmp, flag;
 
	for (i = (int)a; i <= (int)z; i++)
	{
		flag = 0;
 
		if (i == (int)a || i == (int)b || i == (int)l || i == (int)e)
			flag++;
 
		*(rr) = (char)i;
 
		if (flag == 1)
			printf("%c\n", i);
 
		for (j = i + 1; j <= (int)z; j++)
		{
			if (j == (int)a || j == (int)b || j == (int)l || j == (int)e)
				flag++;
 
			if (flag > 1)
			{
				break;
				*(rr + j - i + 1) = '\0';
			}
 
			*(rr + j - i) = (char)j;
 
			if (flag == 1)
			{
				*(rr + j - i + 1) = '\0';
				printf("%s\n", rr);
			}
		}
	}
}

※ 在字符串赋值的过程中，最后需要添加 '\0'，否则会乱码。

collatzeFib.c

1. Write a C-function that takes a positive integer n as argument and prints a series of numbers generated by the following algorithm, until 1 is reached:

   • if n is even, then n ← n/2

   • if n is odd, then n ← 3*n+1

   (Before you start programming, calculate yourself the series corresponding to n=3.)

2. The Fibonacci numbers are defined as follows:
   • Fib(1) = 1
   • Fib(2) = 1

   • Fib(n) = Fib(n-1)+Fib(n-2) for n≥3

   Write a C program that calls the function in Part a. with the first 10 Fibonacci numbers. The program should print the Fibonacci number followed by its corresponding series. The first 4 lines of the output is as follows:

    Fib[1] = 1:
    Fib[2] = 1:
    Fib[3] = 2: 1
    Fib[4] = 3: 10 5 16 8 4 2 1 

a - code

void even_odd(int n)
{
	if (n < 0)
		printf("Please input a positive integer!!!\n");
	else
		printf("%d\n", n);
 
	while (n != 1)
	{
		if (n % 2 == 0)
			n = n / 2;
		else
			n = 3 * n + 1;
		printf("%d\n", n);
	}
}

b - code

int* Fib()
{
	static int ff[10];
	ff[0] = 1;
	ff[1] = 1;
	int i;
	for (i = 2; i < 10; i++)
		ff[i] = ff[i - 2] + ff[i - 1];
	return ff;
}

※ 注意返回数组的方法，另外需要通过 static 关键字来定义数组。

调用数据：

#include <stdio.h>
int main()
{
	void even_odd(int n);
	int* Fib();
 
	int i;
	int* fib_arr;
	fib_arr = Fib();
 
	for (i = 0; i < 10; i++)
	{
		printf("Fib[%d] = %d:", i + 1, fib_arr[i]);
		even_odd(fib_arr[i]);
	}
 
	return 0;
}

参考：C 从函数返回数组

fastMax.c

Write a C-function that takes 3 integers as arguments and returns the largest of them. The following restrictions apply:

• You are permitted to only use assignment statements, a return statement and Boolean expressions

• You are not permitted to use if-statements, loops (e.g. a while-statement), function calls or any data or control structures

int largest(int a, int b, int c)
{
	int max;
	max = (a > b) ? a: b;
	max = (max > c) ? max : c;
	return max;
}