计算机二级考试备考

素颜马尾好姑娘i 2022-12-03 15:50 210阅读 0赞

程序设计题  
【题一】  
![在这里插入图片描述][2020090410554230.png_pic_center]

#include<stdio.h>
    #define N 80
    void fun(int *w,int p,int n)
    {
        
    	int x, j, ch;
    	for (x = 0; x <= p; x++)
    	{
        
    		ch = w[0];
    		for (j = 1; j < n; j++)
    		{
        
    			w[j - 1] = w[j];
    		}
    		w[n - 1] = ch;
    	}
    }
    int main()
    {
        
    	int a[N] = {
         1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 };
    	int i, p, n = 15;
    	printf("The oriqinal data:\n");
    	for (i = 0; i < n; i++)
    	{
        
    		printf("%3d", a[i]);
    	}
    	printf("\n\nEnter p:");
    	scanf("%d", &p);
    	fun(a, p, n);
    	printf("\nThe data after moving:\n");
    	for (i = 0; i < n; i++)
    	{
        
    		printf("%3d", a[i]);
    	}
    	printf("\n\n");
    }

【题二】  
![在这里插入图片描述][20200904105814744.png_pic_center]

void fun(char* a)
    {
        
    	while (*a != '\0')
    	{
        
    		a++;
    	}
    	a--;
    	while (*a == '*')
    	{
        
    		a--;
    	}
    	*(a + 1) = '\0';
    }

【题三】  
![在这里插入图片描述][20200904112236507.png_pic_center]

void fun(int m,int k,int xx[])
    {
        
    	int i, j, n;
    	for (i = m + 1, n = 0; n < k; i++)
    	{
        
    		for (j = 2; j < i; j++)//判断i是否是素数
    		{
        
    			if (i%j == 0)break;//不是跳出判断，判断下一个数
    		}
    		if (j >= i)//若是素数，存入xx[]数组中
    		{
        
    			xx[n++] = i;
    		}
    	}
    }

【题四】  
![在这里插入图片描述][20200912215819660.png_pic_center]

#include<stdio.h>
    int fun(char *s)
    {
        
    	int i, j=0;
    	for (i = 0; s[i] != '\0'; i++)
    	{
        
    		if (s[i] != ' '&&s[i + 1] == ' ' || s[i + 1] == '\0')
    			j++;
    	}
    	return j;
    }
    int main()
    {
        
    	int a;
    	char *s = "i love you";
    	a = fun(s);
    	printf("%d", a);
    	return 0;
    }

【题五】  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2dldHNfcw_size_16_color_FFFFFF_t_70_pic_center]

#include<stdio.h>
    #define N 3
    void fun(int a[][N],int m)
    {
        
    	for (int i = 0; i < N; i++)
    	{
        
    		for (int j = i; j < N; j++)
    		{
        
    			a[i][j] = a[i][j] * m;
    		}
    	}
    }
    int main()
    {
        
    	int a[][N] = {
         {
        1,9,7},{
        2,3,8},{
        4,5,6} };
    	int m = 2;
    	fun(a, m);
    	for (int i = 0; i < N; i++)
    	{
        
    		for (int j = 0; j < N; j++)
    		{
        
    			printf("%d ", a[i][j]);
    			if (j == N - 1)printf("\n");
    		}
    	}
    	return 0;
    }

【题六】  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2dldHNfcw_size_16_color_FFFFFF_t_70_pic_center 1]

#include<stdio.h>
    #define N 5
    float fun(int a[][N])
    {
        
    	float sum = 0;
    	int k = 0;
    	for (int i = 0; i < N; i++)
    	{
        
    		if (i == 0)  //处理第一行
    		{
        
    			for (int j = 0; j < N; j++)
    			{
        
    				sum = sum + a[i][j];
    				k++;
    			}
    		}
    		else if (i == N - 1)  //处理最后一行
    		{
        
    			for (int j = 0; j < N; j++)
    			{
        
    				sum = sum + a[i][j];
    				k++;
    			}
    		}
    		else    //处理中间的行
    		{
        
    			sum = sum + a[i][0] + a[i][N - 1];
    			k = k + 2;
    		}
    	}
    	return sum / k;
    }
    int main()
    {
        
    	float prem;
    	int a[][N] = {
         {
        0,1,2,7,9},{
        1,9,7,4,5},{
        2,3,8,3,1},{
        4,5,6,8,2},{
        5,9,1,4,1} };
    	prem = fun(a);
    	printf("%.3f", prem);
    	return 0;
    }

【题七】  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2dldHNfcw_size_16_color_FFFFFF_t_70_pic_center 2]

#include<stdio.h>
    #define M 4
    #define N 5
    int fun(int a[M][N])
    {
        
    	int sum = 0;
    	for (int i = 0; i < M; i++)
    	{
        
    		if (i == 0)  //处理第一行
    		{
        
    			for (int j = 0; j < N; j++)
    			{
        
    				sum = sum + a[i][j];
    			}
    		}
    		else if (i == M - 1)  //处理最后一行
    		{
        
    			for (int j = 0; j < N; j++)
    			{
        
    				sum = sum + a[i][j];
    			}
    		}
    		else sum = sum + a[i][0] + a[i][N - 1];//处理中间的行
    	}
    	return sum;
    }
    int main()
    {
        
    	int prem;
    	int a[M][N] = {
         {
        1,3,5,7,9},{
        2,9,9,9,4},{
        6,9,9,9,8},{
        1,3,5,7,0}};
    	prem = fun(a);
    	printf("%d", prem);
    	return 0;
    }

【题八】  
![在这里插入图片描述][20200920173548158.png_pic_center]

#include<stdio.h>
    float fun(float h)
    {
        
    	int tmp = (int)(h * 1000 + 5) / 10;
    	return (float)tmp / 100;
    }
    
    int main()
    {
        
    	float h = 1234.564;
    	printf("%7.2f",fun(h));
        return 0;
    }

【题九】  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2dldHNfcw_size_16_color_FFFFFF_t_70_pic_center 3]

#include<stdio.h>
    void fun(int array[3][3])
    {
        
    	int i, j, t;
    	for (i = 0; i < 3; i++)
    	{
        
    		for (j = i + 1; j < 3; j++)
    		{
        
    			t = array[i][j];
    			array[i][j] = array[j][i];
    			array[j][i] = t;
    		}
    	}
    }
    
    int main()
    {
        
    	int array[3][3] = {
         {
        100,200,300},{
        400,500,600},{
        700,800,900} };
    	fun(array);
    	for (int i = 0; i < 3; i++)
    	{
        
    		for (int j = 0; j < 3; j++)
    		{
        
    			printf("%d ", array[i][j]);
    			if (j == 2)printf("\n");
    		}
    	}
        return 0;
    }

【题十】  
![在这里插入图片描述][20200920180734766.png_pic_center]

#include<stdio.h>
    #define MAX 100
    int fun(int lim, int aa[MAX])
    {
        
    	int i, j, k = 0;
    	for (i = 2; i <= lim; i++)
    	{
        
    		for (j = 2; j < i; j++)
    		{
        
    			if (i%j == 0)break;	
    		}
    		if (j >= i)aa[k++] = i;
    	}
    	return k;
    }
    
    int main()
    {
        
    	int sum,lim = 20;
    	int aa[MAX];
    	sum = fun(lim, aa);
    	printf("%d\n", sum);
    	for (int i = 0; i < sum; i++)
    	{
        
    		printf("%d ", aa[i]);
    	}
    	return 0;
    }

【题11】  
![在这里插入图片描述][20200920182713363.png_pic_center]

char *fun(char(*a)[81], int num, char *max)
    {
        
    	int i = 0;
    	max = a[0];
    	for (i = 0; i < num; i++)
    	{
        
    		if (strlen(max) < strlen(a[i]))
    			max = a[i];
    	}
    	return max;
    }

[2020090410554230.png_pic_center]: /images/20221123/328003f2a00d4f7c81732f6dc29c1985.png
[20200904105814744.png_pic_center]: /images/20221123/9f3f75fcc0234320951bced8af88a9e7.png
[20200904112236507.png_pic_center]: /images/20221123/3308ec46c0014acd94787916ccd4466e.png
[20200912215819660.png_pic_center]: /images/20221123/56bebd6291c14658ad634cfc0a9a3ec7.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2dldHNfcw_size_16_color_FFFFFF_t_70_pic_center]: /images/20221123/6ddeac7ba4e7461eb68157e494d52e8d.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2dldHNfcw_size_16_color_FFFFFF_t_70_pic_center 1]: /images/20221123/67e9d18770f54879bb14ca7a9ca67476.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2dldHNfcw_size_16_color_FFFFFF_t_70_pic_center 2]: /images/20221123/e8b7cb972155441da6de0fafdf9c63ca.png
[20200920173548158.png_pic_center]: /images/20221123/190a2a65b29549b1aa3902685fa9bd2a.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2dldHNfcw_size_16_color_FFFFFF_t_70_pic_center 3]: /images/20221123/fd5fe14ef541416e812d2b8c70875e59.png
[20200920180734766.png_pic_center]: /images/20221123/d3e97065beb343758b29de44ccab080f.png
[20200920182713363.png_pic_center]: /images/20221123/a0fc3653daea4fbb97322245639ca430.png