实验4

实验任务1

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

问题1

是；
相同

问题2

是；
相同；
16；
x[0]的值与&x[0][0]相同，x[1]的值与&x[1][0]相同，说明二维数组中的每一个数组的值都是该数组中的第一个数的地址

实验任务2

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("%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("%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

问题1

形参：int x[]；
实参：x

问题2

函数input:输入n个数并赋值到数组x中；
函数compute:对数组中的数去掉一个最高值，去掉一个最低值，再取平均值

实验任务3

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("%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

问题1

形参：int x[][N]；
实参：x

问题2

不能，会报错

问题3

函数output:输出数组中的值，每输出一个数组之后换行；
函数init:对数组赋值

实验任务4

代码

#include <stdio.h>
#define N 100

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

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

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

    return 0;
}

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

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

    return 0;
}

double median(int x[N], int n)
{
    double ans;
    int i,j,temp;

    for (i = 0; i < n; ++i)
    {
        for (j = i + 1; j < n; ++j)
        {
            if (x[i] > x[j])
            {
                temp = x[i];
                x[i] = x[j];
                x[j] = temp;
            }
        }
    }

    if (n % 2 == 0)
        ans = (x[n / 2] + x[n / 2 - 1]) / 2.0;
    else
        ans = x[(n - 1) / 2];

    return ans;
}

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实验任务5

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("输入n: ");
    scanf("%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("%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, j;
    int temp;

    for (i = 0; i < n; ++i)
    {
        temp = x[i][n-1];

        for (j = n-1; j > 0; --j)
            x[i][j] = x[i][j - 1];

        x[i][j] = temp;
    }
}

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实验任务6

代码

#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("%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 s[N];
    int i = 0,j;

    while (x != 0)
    {
        s[i] = x % n;
        x = x / n;
        ++i;
    }

    for (j = i - 1; j >= 0; --j)
    {
        if (s[j] >= 10)
            printf("%c", 'A' + s[j] - 10);  
        else
            printf("%d", s[j]);
    }

    printf("\n");
}

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实验任务7

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("%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("%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 line[N] = { 0 }, column[N] = { 0 }, diagonal[2] = { 0 };
    int l = 1, c = 1, d = 1;
    int i, j;

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

    for (i = 0; i < n-1; ++i)
    {
        if (line[i] == line[i + 1])
            l = 1;
        else
        {
            l = 0;
            break;
        }
    }

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

    for (j = 0; j < n - 1; ++j)
    {
        if (line[i] == line[i + 1])
            l = 1;
        else
        {
            c = 0;
            break;
        }
    }

    diagonal[0] = 0;
    diagonal[1] = 0;

    for (i = 0, j = 0; i < n&& j < n; ++i, ++j)
        diagonal[0] += x[i][j];

    for(i=0,j=n-1;i<n&&j>=0;++i,--j)
        diagonal[1] += x[i][j];

    if (diagonal[0] == diagonal[1])
        d = 1;
    else
        d = 0;

    if (l==1 && c==1 && d==1 && line[0]==column[0] && line[0] == diagonal[0])
        return 1;
    else
        return 0;
}

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实验任务8

代码

#include<stdio.h>
#define N 100

int main()
{
	int x[N];
	int n;
	int square,cube;
	int i,j,k;
	int a;

	for (n = 0;; ++n)
	{
		i = 0;
		a = 1;

		square = n * n;
		cube = n * n * n;

		do{
			x[i] = square % 10;
			square /= 10;
			++i;
		} while (square != 0);

		do{
			x[i] = cube % 10;
			cube /= 10;
			++i;
		} while (cube != 0);

		if (i != 10)
			continue;

		for (j = 0; j < i-1 && a == 1; ++j)
		{
			for (k = j + 1; k < i; ++k)
			{
				if (x[j] == x[k])
				{
					a = 0;
					break;
				}
			}
		}

		if (i == 10 && a == 1)
			break;
	}

	printf("%d", n);

	return 0;
}

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