实验4

任务1

#include<stdio.h>
const int N = 4;

int main()
{
    int a[N] = {2, 0, 2, 1};
    char b[N] = {'2', '0', '1', '1'};
    int i;
    
    printf("sizeof(int) = %d\n", sizeof(int));
    printf("sizeof(char) = %d\n", sizeof(char));
    printf("\n");
    
    for(i=0; i<N; ++i)
        printf("%x: %d\n", &a[i], a[i]);
        
    printf("\n");
    
    for(i=0; i<N; ++i)
        printf("%x: %c\n", &b[i], b[i]);
        
    return 0;
}

连续存放，int型中每个元素占4个内存字节单元，char型占1个

#include<stdio.h>

int main()
{
    int a[2][3] = {{1, 2, 3}, {4, 5, 6}};
    char b[2][3] = {{'1', '2', '3'}, {'4', '5', '6'}};
    int i,j;
    
    for(i=0; i<2; ++i)
        for(j=0; j<3; ++j)
            printf("%x: %d\n", &a[i][j], a[i][j]);
            
    printf("\n");
    
    for(i=0; i<2; ++i)
        for(j=0; j<3; ++j)
            printf("%x: %c\n", &b[i][j], b[i][j]);
}

 连续存放，int型中每个元素占4个内存字节单元，char型占1个

 

任务2

#include<stdio.h>

#define N 1000
int fun(int n, int m, int bb[N])
{
    int i, j, k = 0, flag;
    
    for(j=n; j<=m; j++)
    {
        flag=1;
        for(i=2; i<j; i++)
           if(j%i==0)
           {
                   flag = 0;
                   break;
           }
           if (flag==1)
            {
                bb[k++] = j;
            }   
    }
    
    return k;
}

int main()
{
    int n = 0, m = 0, i, j, k, bb[N];
    
    scanf("%d", &n);
    scanf("%d", &m);
    
    for(i=0; i<m-n; i++)
        bb[i] = 0;
        
    k = fun(n,m,bb);
    
    for(i=0; i<k; i++)
        printf("%4d", bb[i]);
        
    return 0;
}

 

 任务3

#include<stdio.h>
const int N = 5;

int find_max(int x[], int n);
void input(int x[], int n);
void output(int x[], int n);

int main()
{
    int a[N];
    int max;
    
    input(a, N);
    output(a, N);
    max = find_max(a, N);
    
    printf("max = %d\n", max);
    return 0;
}

void input(int x[], int n)
{
    int i;
    
    for (i=0; i<n; ++i)
        scanf("%d", &x[i]);
}

void output(int x[], int n)
{
    int i;
    
    for (i=0; i<n; ++i)
        printf("%d ", x[i]);
    printf("\n");
}

int find_max(int x[], int n)
{
    int i, max=0;
    
    for(i=0;i<n;i++)
    {
        if(x[i]>max)
            max=x[i];
        else
            max=max;
    }
    
    return max;
}

 任务4

#include<stdio.h>
void dec2n(int x, int n);

int main()
{
    int x;
    
    printf("输入一个十进制整数：");
    scanf("%d", &x);
    
    dec2n(x, 2);
    dec2n(x, 8);
    dec2n(x, 16);
    
    return 0;
}

void dec2n(int x, int n)
{
    int i, u[1000];
    char p[1000];
    
    for(i=0;x>0;i++)
    {    
        u[i] = x%n;
        x = x/n;
    
        if(u[i]>=10)
        {
            switch(u[i])
            {
                case 10:p[i]='a';
                case 11:p[i]='b';
                case 13:p[i]='d';
                 case 14:p[i]='e';
                case 15:p[i]='f';
            }
    }
    else
        p[i] = u[i]+'0';
    }
    
    for(i=i-1;i>=0;i--)
        printf("%c",p[i]);

    printf("\n");
}

 

 任务5

#include<stdio.h>
void prty();

int main()
{
    int n,i,j;
    
    printf("Enter n：");
    
    while(scanf("%d", &n)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                if(i<j)
                    printf("%d ",i);
                else
                    printf("%d ",j);
            }
            printf("\n");
        }
        printf("\n");
        printf("Enter n：");
    }
    
    return 0;
}