求代做个C语言编程

挺简单的啊。

测试系统

列出10个数学题

输入每道题的答案。

如果用户的答案不正确，显示出正确答案。

给出最终的分数。

用switch语句来控制加或减的运算。

代码稍后~

#include <stdio.h>

#include <time.h>

#include <stdlib.h>

#define ADD 0

#define MINUS 1

void main()

{

int i,a,b,answer,score=0,op,rtanswer

srand((unsigned)time(NULL))

for (i=0i<10i++)

{

a=100*rand()/RAND_MAX

b=100*rand()/RAND_MAX

op=2*rand()/RAND_MAX

switch (op)

{

case ADD:

printf("%d %d+%d=",i+1,a,b)

rtanswer=a+b

break

case MINUS:

printf("%d %d-%d=",i+1,a,b)

rtanswer=a-b

break

}

scanf("%d",&answer)

if (answer!=rtanswer)

{

printf("correct answer=%d\n",rtanswer)

}

else score++

}

printf("Your Score is %d\n",score)

}

////////////////////////////////第一题

/*

Question 1 - 15 marks

Write a C program that prints out all dierent possibilities of obtaining $2(2

pounds) using coins of values 2 pence, 5 pence, and 10 pence. Indicate how

many possibilities have been found. The output of your program may look like:

$2 = 100 x 2p

$2 = 90 x 2p + 4 x 5p

$2 = 80 x 2p + 8 x 5p ....

In total, there are ??? possibilities to make $2.

Note that ??? (no. of the possibilities) is calculated by your C program.

*/

#include<stdio.h>

void main()

{

int count2,count5,count10

int count=0

int sum

count=0

for( count2=100count2>=0 count2=count2-1 )

for( count5=40count5>=0count5=count5-1 )

for( count10=20count10>=0count10=count10-1 )

{

sum=count2*2 + count5*5 + count10*10

if( sum==200 )

{

printf("$2 =")

if( count2!=0 )

printf(" +%d X 2p",count2)

if( count5!=0 )

printf(" +%d X 5p",count5)

if( count10!=0 )

printf(" +%d X 10p",count10)

printf("\n")

count++

}

}

printf("In total, there are %d possibilities to make $2.\n",count)

}

///////////////////////////////////////////第二题

/*

Question 2 - 15 marks

Write a C program that reads m x n matrix "A" and p x q matrix "B", checks

whether these matrices are multipliable in either order or not (e.g. whether A

x B or B x A is dened). Further, if A x B or B x A is dened then calculates

the product and prints out the result.

*/

/* 矩阵元素是整型的 */

#include<stdio.h>

#define MAX 256

void SetZero( int a[MAX][MAX] )

void Print( int a[MAX][MAX], int r, int c )

void MultiplyMatrix( int a[MAX][MAX],int m,int n, int b[MAX][MAX],int p, int q )

int result[MAX][MAX]//存放结果

/* 矩阵相乘 */

void MultiplyMatrix( int a[MAX][MAX],int m,int n, int b[MAX][MAX],int p, int q )

{

int row,column

SetZero(result)

if( n==p )

{

for( row=0row<mrow++ )

for( column=0column<qcolumn++ )

for( n=0n<pn++ )

result[row][column] += a[row][n] * b[n][column]

}

else if( q==m )

{

for( row=0row<prow++ )

for( column=0column<ncolumn++ )

for( m=0m<qm++ )

result[row][column] += b[row][m] * a[m][column]

}

else

printf("can't multiply.\n")

}

/* 输出矩阵 */

void Print( int a[MAX][MAX], int r, int c )

{

int i,j

for( i=0i<ri++ )

{

for( j=0j<cj++ )

printf("%4d",a[i][j])

printf("\n")

}

}

/*将矩阵元素设为0 */

void SetZero( int a[MAX][MAX] )

{

int i,j

for( i=0i<MAXi++ )

for( j=0j<MAXj++ )

a[i][j]=0

}

void main()

{

int a[MAX][MAX]

int b[MAX][MAX]

int m,n//a[m][n]

int p,q//b[p][q]

int i,j

char f='A'// f=='L' A*B

// F=='R' B*A

SetZero(a)

SetZero(b)

/* 输入矩阵A的元素,需先输入m和n(矩阵的行列) */

printf("Input m*n matrix \"A\" \n")

printf("m=")scanf("%d",&m)

printf("n=")scanf("%d",&n)

printf("matrixA:\n")

for( i=0i<mi++ )

for( j=0j<nj++ )

scanf("%d",&a[i][j])

/* 输入矩阵B的元素,需先输入p和q(矩阵的行列) */

printf("Input p*q matrix \"B\" \n")

printf("p=")scanf("%d",&p)

printf("q=")scanf("%d",&q)

printf("matrixB:\n")

for( i=0i<pi++ )

for( j=0j<qj++ )

scanf("%d",&b[i][j])

if( n==p )

{

MultiplyMatrix( a,m,n, b,p,q )

f='L'

}

else if( q==m )

{

MultiplyMatrix( b,p,q, a,m,n )

f='R'

}

else

{

printf("can't multiply.\n")

f='A'

}

printf(" matrix A\n")

Print( a, m, n )

printf(" matrix B\n")

Print( b, p, q )

if( f=='L' )

{

printf(" A*B\n")

Print( result, m, q )

}

else if( f=='R' )

{

printf(" B*A\n")

Print( result, p, n )

}

}

//////////////////////////////////第三题

/*

Question 3 - 20 marks

Write a C program that initializes an array of integer and then copies the con-

tents of the array into two other arrays. Declare all arrays in the main program.

To make the rst copy, write a function, which uses the array notation (the

square brackets []) to access the elements of the array.

To make the second copy, write a function that uses the pointer notation and

pointer incrementing to access the elements of the arrays.

Each function takes the name of the source array, the name of the target/destination

1

array and the number of elements to be copied as function arguments.

Here is an example showing how the functions should be called giving the fol-

lowing declarations:

int source[4] = {1,2,4,6}

int destination1[4]

int destination2[4]

copy_array(source, destination1, 4)

copy_ptr(source, destination2, 4)

*/

#include<stdio.h>

void Print( int s[], int n )

{

int i

for( i=0i<ni++ )

printf("%4d",s[i])

printf("\n")

}

void copy_ptr( int *s, int *d, int n )

{

for( n--n>=0n-- )

{

*d=*s

d++

s++

}

}

void copy_array( int s[], int d[], int n )

{

int i

for( i=0i<ni++ )

d[i]=s[i]

}

void main()

{

int source[4] = {1,2,4,6}

int destination1[4]

int destination2[4]

copy_array(source, destination1, 4)

copy_ptr(source, destination2, 4)

Print( source, 4 )

printf("source[] = ")

Print( source, 4 )

printf("destination1[] = ")

Print( destination1, 4 )

printf("destination2[] = ")

Print( destination2, 4 )

}

////////////////////////////////第四题

晚上我再写第四题吧 有点长

两种方法我写在一起，可以独立拆开。

#include <stdio.h>

void finda1(char a[3][10])

void finda2(char a[3][10])

void show(char (*p)[10])

int main()

{

  char a[3][10]={{"gehajl"},{"788a987a7"},{"ccabbbabbb"}}

  printf("原数组内容:\n")

  show(a)

  printf("\n1、用数组指针的方法（函数finda1）：\n")

  finda1(a)

  printf("执行后:\n")

  show(a)

  printf("\n---------------------\n")

  char b[3][10]={{"gehajl"},{"788a987a7"},{"ccabbbabbb"}}

  printf("原数组内容:\n")

  show(a)

  printf("\n2、用指针数组的方法（函数finda2）：\n")

  finda2(b)

  printf("执行后:\n")

  show(b)

  return 0

}

void finda1(char a[3][10])

{

  int i,j

  char (*p)[10]=a

  for(i=0i<3i++)

      for(j=0j<10j++)

          if(p[i][j]=='a')

              printf("发现：第%d行第%d个元素是‘a’，已替换\n",i+1,j+1),p[i][j]='1'

}

void finda2(char a[3][10])

{

  int i,j

  char *p[3]={&a[0][0],&a[1][0],&a[2][0]}

  for(i=0i<3i++)

      for(j=0j<10j++)

          if(p[i][j]=='a')

              printf("发现：第%d行第%d个元素是‘a’，已替换\n",i+1,j+1),p[i][j]='1'

}

void show(char (*p)[10])

{

  int i,j

  for(i=0i<3i++,printf("\n"))

      for(j=0j<10j++)

          printf("%c ",p[i][j])

}

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