❤️学弟学妹要的❤️大学生必备C语言基础练习题目来了（四）（建议收藏哦）❤️

Love The Way You Lie 2022-09-11 03:18 58阅读 0赞

**题目1. 编程求x的y次方，fun函数实现x的y次方的计算，main函数中调用fun函数，计算4的3次方的值。请画出函数fun的流程图。**

**流程图：**

![image-20210911162040845][]

**实现代码：**

#include<stdio.h>
    #include<math.h>
    void main()
    { 
     //定义fun函数
     double fun(int x, int y);
     int x, y;
     double z;
     printf("请分别输入x,y的值：");
     scanf("%d %d",&x,&y);
     z = fun(x,y);      //调用fun函数
     printf("%d的%d次方 = %lf\n" , x,y,z);
    
    }
    
    double fun(int x, int y)
    { 
       double pow(double x, double y);
       double m = 0;
       m = pow(x,y);   //求次方函数
       return m;
    }

**运行结果：**  
![image-20210911163144834][]

--------------------

**题目2． 编写一个判断素数的函数fun，在主函数中计算1000以内所有素数的平均值并输出。请画出判断素数函数的流程图。**

**流程图：**

![image-20210911162214305][]

**实现代码：**

#include<stdio.h>
    int fun(int x)
    { 
        int i;
        for (i=2;i<x;i++)
        { 
          if (x%i == 0)
          { 
           return 0;
          }
        }
           return 1;
    }
    void main()
    { 
       int i;
       float avg = 0,count = 0,sum = 0;
       for(i=2;i<=1000;i++)
       { 
         if(fun(i))
    	 { 
    	   count ++;
    	   sum += i;
    	 }  
       }
           avg = sum / count;
           printf("平均数为：%lf\n",avg);
    }

**运行结果：**  
![image-20210911163154787][]

--------------------

**题目3．编写四个函数，分别计算N*N矩阵转置、平均值、对角线元素之和、周边元素之和。在主程序中调用这四个函数并输出结果。N*N矩阵的数值从键盘输入。请画出求N\*N矩阵转置函数的流程图。**

**流程图：**

![image-20210911162242797][]  
**实现代码：**

#include<stdio.h>
    #define N 3
    //转置函数
    int transp(int a[N][N])
    { 
    	int i,j,b[N][N];
    	for(i = 0; i < N; i++)
    	{ 
    		for(j = 0; j < N; j++)
    		{ 
    			b[j][i]=a[i][j];
    		}
    	}
    	for(i = 0; i < N; i++)
    	{ 
    		for(j = 0; j < N; j++){ 
    		   printf("%3d ",b[i][j]);
    		}
    		 printf("\n");
    	}
    	return 0;
    }
    //平均值
    double average(int a[N][N])
    { 
    	int i,j;
    	double avg,sum=0,count=0;
    	for(i = 0; i < N; i++)
    	{ 
    		for(j = 0; j < N; j++)
    		{ 
    			sum += a[i][j];
    			count++;
    		}
    	}
    	avg = sum / count;
    	return avg;
    }
    
    //对角线元素之和
    int diagonal_sum(int a[N][N])
    { 
    	int i,j, s = 0;
    	for(i = 0; i < N; i++)
    	{ 
    		for(j = 0; j < N; j++)
    		{ 
    			if(i == j)
    			{ 
    			  s += a[i][j];
    			}
    		}
    		s += a[i][N-i-1];
    	}
    	if(N % 2 == 1)
    	{ 
    	  s -= a[N/2][N/2];
    	}
    	return s;
    }
    
    //周边元素之和
    int round_sum(int a[N][N])
    { 
    	int i, s = 0;
    	for (i = 0;i < N;i++)
    		s += a[0][i] + a[N - 1][i]; //第一行和最后一行的数据
    	for (i = 1;i < N - 1;i++)      
    		s += a[i][0] + a[i][N - 1]; //第一列的第二个数开始和最后一列的第二个数开始
    	return s;
    }
    
    void main()
    { 
    	int a[N][N];
    	int i,j;
    	int sum1,sum2;
    	double avg1;
    	for(i = 0; i < N; i++){ 
    		for(j = 0; j < N; j++){ 
    		   scanf("%d",&a[i][j]);
    		}
    	}
    	sum1 = round_sum(a); 
    	printf("周边元素之和: %d\n",sum1);
    	sum2 = diagonal_sum(a); 
    	printf("对角线元素之和: %d\n",sum2);
    	avg1 = average(a); 
    	printf("平均数: %lf\n",avg1);
    	printf("转置矩阵: \n");
    	transp(a);
    }

**运行结果：**  
![image-20210911163205179][]

--------------------

**题目4. 编写一个函数，使用递归方法求阶乘n!。在主程序中调用这个函数并输出结果。n从键盘输入。请画出递归函数的流程图。**

**流程图：**

![image-20210911162341598][]  
**实现代码：**

#include<stdio.h>
    int recursion(int n)
    { 
       if (n <0)
       { 
          printf("n<0,输入错误请输入大于0的数 \n");
       }
       else if(n==0 || n==1)
       { 
          return 1;
       }
       else
       { 
          return n*recursion(n - 1);
       }
    }
    
    void main()
    { 
       int n,result=0;
       printf("请输入一个正整数: ");
       scanf("%d", &n);
       result = recursion(n);
       printf("result=%d\n",result);
    }

**运行结果：**  
![image-20210911163212017][]

--------------------

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[image-20210911162040845]: /images/20220828/0196ac44639242d0b9aaf210c6b9e58f.png
[image-20210911163144834]: /images/20220828/45080f5ffa754b138611fed634f13bef.png
[image-20210911162214305]: /images/20220828/eaee21d2f60e478aae1e9e74de322159.png
[image-20210911163154787]: /images/20220828/469a468cfc1f4ae98dbe95e750ba78d0.png
[image-20210911162242797]: /images/20220828/1c7a8491a642470282760d469451e1c7.png
[image-20210911163205179]: /images/20220828/fa83a1d9736641ff925e5ab1e9770588.png
[image-20210911162341598]: /images/20220828/9be9a38f60684e568423169dd2fa3a32.png
[image-20210911163212017]: /images/20220828/3b8a4e6984e2490d9c906b3678e55573.png
[26e793d93499cb2c2afa75ea987e42a3.png]: /images/20220828/8ff7c3184eeb4245815a8ba295e15ca2.png