实验4

task1

#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;
}

int型数组a在内存中连续存放,每个元素占用4个字节单元

char型数组b在内存中连续存放,每个元素占用1个字节单元

#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]);
}

 

 int型二维数组a在内存中不连续存放,每个元素占用4个字节单元

char型二维数组b在内存中不连续存放,每个元素占用1个字节单元

 

 

 

 

 

task2

#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!=0)
            bb[k++] = j;
    }
    return k;
}
int main()
{
    int n = 0, m = 0, i, 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;
}

 

 

 

 

 

 

task3

#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 temp;
    int i, j;
    for (i = 0; i < n-1; i++)
    {
        for (j = 0; j < n - i - 1; j++)
        {
            if (x[j] < x[j + 1])
            {
                temp = x[j];
                x[j] = x[j + 1];
                x[j + 1] = temp;
            }
        }
    }
    return x[0];
}

 

 

 

 

 

task4

#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 a[32],i=0,b[32];
    while (x > 0)
    {
        a[i] = x % n;
        i = i + 1;
        x = x / n;
    }
    if (n == 16)
    {
        for (; i > 0; i--)
        printf("%d", a[i - 1]);
    }
    else    
    for (; i > 0; i--)
        printf("%d", a[i - 1]);
    printf("\n");
}

 

 

 

 

 

task5

#include <stdio.h>
#define N 100
void jz(int a[N][N], int n);
int main()
{
    int i, j, n,a[N][N];
    printf("Enter n:");
    while (scanf("%d", &n) != EOF)
    {
        jz(a, n);
        for (i = 1; i <= n; i++)
        {
            for (j = 1; j <= n; j++)
                printf("%d", a[i][j]);
            printf("\n");
        }
    printf("Enter n:");
    }
    return 0;
}
void jz(int a[N][N], int n)
{
    int i, j;
    for (i = 1; i <= n; i++)
    {
        for (j = 1; j <= n; j++)
        {
            if (i < j)
            {
                a[i][j] = i;
            }
            else
            {
                a[i][j] = j;
            }
        }
    }
}

 

posted @ 2021-12-01 20:03  Dayoza  阅读(19)  评论(0编辑  收藏  举报