实验4

task1源代码

#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;
}

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问题1：连续存放，相同

问题2：连续存放，相同，1，第一组和第二组的第一个数的差值

task2源代码

#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;
}

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问题1：数组型和具体数字

问题2：输入数组，输出数组

task3源代码

#include <stdio.h>
#define N 100
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);
        output(x, n); 
        printf("\n");
    }
    return 0;
}
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");
    }
}
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;
}

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问题1：数组和数字

问题2：不能

问题3：输出数组，将数字填入数组

task4源代码

#include <stdio.h>
#define N 100
void input(int x[], int n);
double median(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) {
    int i = 0;
    for (i; i < n; i++)
        scanf_s("%d", &x[i]);
}
double median(int x[], int n) {
    int i, high, low;
    double ans;
    for (i = 0; i < n-1; i++) {
        for (high = 0; high < n-1- i; high++) {
            if (x[high] > x[high + 1]) {
                low = x[high + 1];
                x[high + 1] = x[high];
                x[high] = low;
            }
        }
}
    if (n % 2 == 1) {
        return x[n / 2];
    }
    else {
        return (x[n / 2 - 1] + x[n / 2]) / 2.0;
    }
}

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task5源代码

#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("Enter 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 a[N];
        for (int i = 0; i < n; i++) {
            a[i] = x[i][n - 1];
        }
    for (int j = n - 1; j > 0; j--) {
        for (int i = 0; i < n; i++) {
            x[i][j] = x[i][j - 1];
        }
    }
    for (int i = 0; i < n; i++) {
        x[i][0] = a[i];
    }
}

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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 a, b=0;
    char ans[N];
    while (x > 0) {
        a = x % n;
        if (a < 10)
            ans[b++] = '0' + a;
        if (a >= 10)
            ans[b++] = 'A' + (a - 10);
        x = x / n;
    }
    for (int i = b - 1; i >= 0; i--) {
        printf("%c", ans[i]);
}
    printf("\n");
}

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task7源代码

#include <stdio.h>
#define N 100
void input(int x[][N], int n);
void output(int x[][N], int n);
int is_magic(int x[][N], int n);
int main() {
    int x[N][N];
    int n;
    while (printf("输入n: "), scanf_s("%d", &n) != EOF) {
        printf("输入方阵:\n");
        input(x, n);
        printf("输出方阵:\n");
        output(x, n);
        if (is_magic(x, n))
            printf("是魔方矩阵\n\n");
        else
            printf("不是魔方矩阵\n\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");
    }
}
int is_magic(int x[][N], int n) {
    int i, j;
    int a = n * (n * n + 1) / 2;
    for (i = 0; i < n;i++) {
        int ans = 0;
        for (j = 0; j < n; j++) {
            ans += x[i][j];
        }
        if (ans != a) {
            return 0;
        }
    }
    for (j = 0; j < n; j++) {
        int ans2 = 0;
        for (i = 0; i < n; i++) {
            ans2 = x[i][j];
        }
        if (ans2 != a) {
            return 0;
        }
    }
    int ans3 = 0;
    for (i = 0; i < n; i++) {
        ans3 += x[i][i];
    }
    if (ans3 != a) {
        return 0;
    }
    int ans4 = 0;
    for (i = 0; i < n; i++) {
        ans4 += x[i][n - 1 - i];
    }
    if (ans4 != a) {
        return 0;
    }
    return 1;
}

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