【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.

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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

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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}

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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

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#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?

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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

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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

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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：

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#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

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