实验4

task1

#include <stdio.h>
#define N 4
#define M 2

void test1() {
    int x[N] = { 1, 9, 8, 4 };
    int i;

    printf("sizeof(x) = %d\n", sizeof(x));

    for (i = 0; i < N; ++i)
        printf("%p: %d\n", &x[i], x[i]);

    printf("x = %p\n", x);
}

void test2() {
    int x[M][N] = { {1, 9, 8, 4}, {2, 0, 4, 9} };
    int i, j;

    printf("sizeof(x) = %d\n", sizeof(x));

    for (i = 0; i < M; ++i)
        for (j = 0; j < N; ++j)
            printf("%p: %d\n", &x[i][j], x[i][j]);
    printf("\n");

    printf("x = %p\n", x);
    printf("x[0] = %p\n", x[0]);
    printf("x[1] = %p\n", x[1]);
    printf("\n");
}

int main() {
    printf("测试1: int型一维数组\n");
    test1();

    printf("\n测试2: int型二维数组\n");
    test2();

    return 0;
}

问题1：是，一样

问题2：是，一样。相差16个字节，意义为一行有4个元素

task2

#include <stdio.h>
#define N 100

void input(int x[], int n);
double compute(int x[], int n);

int main() {
    int x[N];
    int n, i;
    double ans;

    while (printf("Enter n: "), scanf_s("%d", &n) != EOF) {
        input(x, n);          
        ans = compute(x, n); 
        printf("ans = %.2f\n\n", ans);
    }

    return 0;
}

void input(int x[], int n) {
    int i;

    for (i = 0; i < n; ++i)
        scanf_s("%d", &x[i]);
}

double compute(int x[], int n) {
    int i, high, low;
    double ans;

    high = low = x[0];
    ans = 0;

    for (i = 0; i < n; ++i) {
        ans += x[i];

        if (x[i] > high)
            high = x[i];
        else if (x[i] < low)
            low = x[i];
    }

    ans = (ans - high - low) / (n - 2);

    return ans;
}

问题：input功能是输入n个整数。compute功能为在数组中数去掉一个最大数和一个最小数，其余数求平均值

task3

#include <stdio.h>
#define N 100

// 函数声明
void output(int x[][N], int n);
void init(int x[][N], int n, int value);

int main() {
    int x[N][N];
    int n, value;

    while (printf("Enter n and value: "), scanf_s("%d%d", &n, &value) != EOF) {
        init(x, n, value);  // 函数调用
        output(x, n);       // 函数调用
        printf("\n");
    }

    return 0;
}

// 函数定义
void output(int x[][N], int n) {
    int i, j;

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

// 函数定义
void init(int x[][N], int n, int value) {
    int i, j;

    for (i = 0; i < n; ++i)
        for (j = 0; j < n; ++j)
            x[i][j] = value;
}

 

问题1：第二维

问题2：output将这个二维数组按n行 n 列输出。init 将数字输入这个二维数组

task4

#include <stdio.h>
#define N 100

double median(int x[], int n);
void input(int x[], int n);

int main() {
    int x[N];
    int n;
    double ans;

    while (printf("Enter n: "), scanf_s("%d", &n) != EOF) {
        input(x, n);
        ans = median(x, n);
        printf("ans = %g\n\n", ans);
    }

    return 0;
}

void input(int x[], int n) {
    for (int i = 0; i < n; i++) {
        scanf_s("%d", &x[i]);
    }
}
double median(int x[],int n) {
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0; j < n - i - 1; j++) {
            if (x[j] > x[j + 1]) {
                int temp = x[j];
                x[j] = x[j + 1];
                x[j + 1] = temp;
            }
        }
    }if (n % 2 == 1)
        return (double)x[n / 2];
    else
        return((double)x[n / 2 - 1] + (double)x[n / 2]) / 2;
}

task5

#include <stdio.h>
#define N 100

void input(int x[][N], int n);
void output(int x[][N], int n);
void rotate_to_right(int x[][N], int n);
int main() {
    int x[N][N];
    int n;

    printf("输入n: ");
    scanf_s("%d", &n);
    input(x, n);

    printf("原始矩阵:\n");
    output(x, n);

    rotate_to_right(x, n);
    printf("变换后矩阵:\n");
    output(x, n);

    return 0;
}

void input(int x[][N], int n) {
    int i, j;

    for (i = 0; i < n; ++i) {
        for (j = 0; j < n; ++j)
            scanf_s("%d", &x[i][j]);
    }
}

void output(int x[][N], int n) {
    int i, j;

    for (i = 0; i < n; ++i) {
        for (j = 0; j < n; ++j)
            printf("%4d", x[i][j]);

        printf("\n");
    }
}
void rotate_to_right(int x[][N], int n) {
    int i, t, j;
    int e[N][N];

    for (i = 0; i < n; ++i)
        for (j = 0; j < n; ++j)
        {
            if (i == n - 1)

                e[j][0] = x[j][i];
            else

                e[j][i + 1] = x[j][i];
        }
    for (i = 0; i < n; ++i)
        for (j = 0; j < n; ++j)
            x[j][i] = e[j][i];
}

 

 

task6

#include <stdio.h>
#define N 100

void dec_to_n(int x, int n); 

int main() {
    int x;

    while (printf("输入十进制整数: "), scanf_s("%d", &x) != EOF) {
        dec_to_n(x, 2); 
        dec_to_n(x, 8);  
        dec_to_n(x, 16); 

        printf("\n");
    }

    return 0;
}

void dec_to_n(int x, int n) {
    int num[N], i = 0;
    if (x == 0) {
        printf("0");
        return;
    }
    while (x > 0) {
        num[i++] = x % n;
        x /= n;
    }
    for (i--; i >= 0; i--) {
        if (num[i] < 10)
            printf("%d", num[i]);
        else
            printf("%c", 'A' + num[i] - 10);
    }
    printf("\n");
}