C语言作业EX06&EX07（修改一次）

  鉴于老师可能没时间讲函数，但是又必须交函数的作业。我把写的代码分享一下，希望能有所帮助。全抄有风险，如果有问题欢迎指正。
  这次作业让我明白一个道理：管它写的有多烂，能跑起来，是对的就行。
  (CPF群里面有人说：取整的话，正数可以直接加0.5取整，但是负数不行，所以为了严谨，加上负数的取整)

Exercise 6: Simple Functions

  1. Write a C program that reads several numbers and uses the function round_to_nearest to round each of these numbers to the nearest integer. The program should print both the original number and the rounded number. 

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    #include <stdio.h>
    int round_to_nearest(float num);
    int main()
    {
    	float num;
        int result;
    	while(scanf("%f",&num)!=EOF)
    	{
    	result = round_to_nearest(num);
    	printf("%d\n",result);
    	}return 0;
    	
    } 

    int round_to_nearest(float num)
    {
    	if (num>=0)                 //正数取整
    	return ((int)(num+0.5));
      	else                        //负数取整
      	return ((int)(num-0.5));    
    }


  2. Write a program that reads three pairs of numbers and adds the larger of the first pair, the larger of the second pair and the larger of the third pair. Use a function to return the larger of each pair.

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    #include<stdio.h>
    float max(float,float);
    int main()
    {
    	int i;
    	float first,second,result,larger,sum=0;
        for(i=0;i<3;i++)
    	{
    	    scanf("%f %f",&first,&second);
    		result = max(first,second);
    		sum+=result;
    	    printf("%f\n",result);
    	}
    	printf("%f",sum);
    	return 0;
    	
    }

    float max(float first,float second){
    	float larger;
    	if(first>=second)
    	larger = first;
    	else
    	larger = second;

    	return larger;

    }

  3. A car park charges a £2.00 minimum fee to park for up to 3 hours, and an additional £0.50 for each hour or part hour in excess of three hours. The maximum charge for any given 24-hour period is £10.00. Assume that no car parks for more than 24 hours at a time.

    Write a C program that will calculate and print the parking charges for each of 3 customers who parked their car in the car park yesterday. The program should accept as input the number of hours that each customer has parked, and output the results in a neat tabular form, along with the total receipts from the three customers:

    Car Hours Charge
    1 1.5 2.00
    2 4.0 2.50
    3 24.0 10.00
    TOTAL 29.5 14.50

    The program should use the function calculate_charges to determine the charge for each customer.

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    #include<stdio.h>
    //这个如果用for的话，我没想到能打印出表格效果的方法。
    float fee(float);
    int main(){
    	int i,money,TOTAL;
    	float fee1,fee2,fee3,time,first=0,second=0,third=0,sum=0,hours=0;
    	printf("The first car's hours is ");
    	scanf("%f",&first);
    	printf("\nThe second car's hours is ");
    	scanf("%f",&second);
    	printf("\nThe third car's hours is ");
    	scanf("%f",&third);
    	fee1 = fee(first);
    	fee2 = fee(second);
    	fee3 = fee(third);
    		time=first+second+third;
            sum=fee1+fee2+fee3;
    	printf ("\nCar	Hours   Charge\n");
    	printf ("  1       %.1lf     %.2lf\n  ",first,fee1);
    	printf ("2       %.1lf     %.2lf\n  ",second,fee2);
    	printf ("3       %.1lf    %.2lf\n  ",third,fee3);
    	printf ("Total   %.2lf   %.2lf",time,sum);
    	
        return 0;
    }

    float fee(float hours){
    	float money;
    	if(hours<=3)
    	money = 2.00;
    	else if(hours>=3&&hours<=19)
    	money = 2 + 0.5*(hours-3);
    	else
    	money = 10;

        return money;	
    }

      这个是用for循环写的。

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    #include<stdio.h>


    //用了For循环的，但是无法打印出表格的效果 
    float fee(float);
    int main(){
    	int i,money,TOTAL;
    	float charge,time,sum=0,hours=0;
    	printf("Car   Hours\n");
    	for(i=0;i<3;i++)
    	{
    		printf("%d\t",i+1);
    	do{
    	scanf("%f",&hours);
    	}while(hours<0&&hours>24);
    		charge = fee(hours);
    		time+=hours;
            sum+=charge;
    		printf("Charge %.2f\n",charge);
    	}
    	printf("Total  %.2lf   %.2lf",time,sum);
        return 0;
    }

    float fee(float hours){
    	float money;
    	if(hours<=3)
    	money = 2.00;
    	else if(hours>=3&&hours<=19)
    	money = 2 + 0.5*(hours-3);
    	else
    	money = 10;

        return money;	
    }

Exercise 7: More Functions

  1. Write a program that uses sentinel-controlled repetition to take an integer as input, and passes it to a function even which uses the modulus operator to determine if the integer is even. The function even should return 1 if the integer is even, and 0 if it is not.

    The program should take the value returned by the function even and use it to print out a message announcing whether or not the integer was even.

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    #include <stdio.h>
    int even(int num);
    int main()
    {
    	int a,num;
        while(scanf("%d",&num)!=EOF)
        {
    	a = even(num);
    	if(a==1)
    	printf("%d is an even\n",num);
    	else if (a==0)
    	printf("%d is an odd\n",num);
        }
        return 0;
    }

    int even(int num){
    	if(num%2)
    	return 0;
    	else
    	return 1;
    	
    }

  2. Write a C program that uses the function integerPower1(base, exponent) to return the value of:

                             baseexponent

    so that, for example, integerPower1(3, 4) gives the value 3 * 3 * 3 * 3. Assume that exponent is a positive, non-zero integer, and base is an integer. The function should use a for loop, and make no calls to any math library functions.

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    #include<stdio.h>
    int intergerPower1(int base,int exponent);
    int main()
    {
    	int base,exponent,result;
    	while(scanf("%d %d",&base,&exponent)!=EOF)
        {
    	    result = intergerPower1(base,exponent);
            printf("%d\n",result);
        }
    	return 0;
    }

    int intergerPower1(int base,int exponent){
    	int i,result1=1;
    	for(i=0;i<exponent;i++)
    	result1*=base;
    	return result1;
    }

  3. Write a C program that uses the recursive function integerPower2(base, exponent) to return the value of:

                         baseexponent

    so that, for example, integerPower2(3, 4) gives the value 3 * 3 * 3 * 3. Assume that exponent is a positive, non-zero integer, and base is an integer. The function should make no calls to any math library functions.
    (Hint: the recursive step will use the relationship:

                         baseexponent = base . baseexponent - 1

    and the base case will be when exponent is 1 since : base1 = base.)

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    #include<stdio.h>
    int intergerPower2(int base,int exponent);
    int main()
    {
    	int base,exponent,result;
    	while(scanf("%d %d",&base,&exponent)!=EOF)
        {
    	    result = intergerPower2(base,exponent);
            printf("%d\n",result);
        }
    	return 0;
    }

    int intergerPower2(int base,int exponent){
        int result2=1;
        if(exponent<=0)
        	return 1;
        return base*intergerPower2(base,exponent-1);
      
    }