实验四

任务一

#include <stdio.h>
#define N 4
int main() {
int a[N] = { 1, 9, 8, 4 };
char b[N] = { '1', '9', '8', '4' };
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("%p: %d\n", &a[i], a[i]);
printf("\n");

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

printf("a = %p\n", a);
printf("b = %p\n", b);
return 0;
}

1：int型数组a，在内存中是连续存放的。每个元素占用4个字节。

2：char型数组b，在内存中是连续存放的。每个元素占用1个字节。

3：是；是。

#include <stdio.h>
#define N 2
#define M 4
int main() {
int a[N][M] = { {1, 9, 8, 4}, {2, 0, 2, 2} };
char b[N][M] = { {'1', '9', '8', '4'}, {'2', '0', '2', '2'} };
int i, j;

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

for (i = 0; i < N; ++i)
for (j = 0; j < M; ++j)
printf("%p: %c\n", &b[i][j], b[i][j]);
return 0;
}

1：是；4。

2：是；

1任务二

#include<stdio.h>
#define N 13

int days_of_year(int, int, int);
int main()
{
int day, month, year;
int days;
while (scanf_s("%d%d%d", &year, &month, &day) != EOF) {
days = days_of_year(year, month, day);
printf("%4d-%02d-%02d是这一年的第%d天.\n\n", year, month, day, days);
}
return 0;
}

int days_of_year(int year, int month, int day)
{
int x[12] = { 31,0,31,30,31,30,31,31,30,31,30,31 };
int leapyear;
leapyear = year % 4 == 0 && year % 100 != 0 || year % 400 == 0;
x[1] = 28 + leapyear;
int s = 0, i;
for (i = 0; i < month - 1; i++)
{
s = s + x[i];
}
return s + day;
}

任务三

 

#include <stdio.h>
#define N 5
void input(int x[], int n);
void output(int x[], int n);
double average(int x[], int n);
void bubble_sort(int x[], int n);
int main() {
int scores[N];
double ave;
printf("录入%d个分数:\n", N);
input(scores, N);
printf("\n输出课程分数: \n");
output(scores, N);
printf("\n课程分数处理: 计算均分、排序...\n");
ave = average(scores, N);
bubble_sort(scores, N);
printf("\n输出课程均分: %.2f\n", ave);
printf("\n输出课程分数(高->低):\n");
output(scores, N);
return 0;
}

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

void output(int x[], int n) {
int i;
for (i = 0; i < n; ++i)
printf("%d ", x[i]);
printf("\n");
}
double average(int x[], int n)
{
int j;
double s = 0.0;
double result;
for (j = 0; j < n; j++)
{
s = s + x[j];
}
result = s / (n * 1.0);
return result;
}
void bubble_sort(int x[], int n)
{
int h, g, t;
for (g = 0; g < n; g++) {
for (h = 0; h < n; h++)
{
if (x[h] < x[h + 1]) {
t = x[h];
x[h] = x[h + 1];
x[h + 1] = t;
}

}
}
}

任务四

 

#include <stdio.h>
#define N 100
void dec2n(int x, int n);
int main() {
int x;
printf("输入一个十进制整数: ");
while (scanf_s("%d", &x) != EOF) {
dec2n(x, 2);
dec2n(x, 8);
dec2n(x, 16);
printf("\n输入一个十进制整数: ");
}
return 0;
}
void dec2n(int x, int n)
{
int i, j = 0;
char y[16] = { '0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F' };
char k[16];
while (x != 0)
{
i = x % n;
x = x / n;
k[j++] = y[i];
}
while (j >= 0)
printf("%c", k[--j]);
printf("\n");
}

任务五

#include <stdio.h>
#define N 100
void func(int x[][N], int n);
void output(int x[][N], int n);
int main() {
int x[N][N];
int i, j, n;
printf("Enter n: ");
while (scanf_s("%d", &n) != EOF) {
func(x, n);
output(x, n);
printf("\nEnter 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("%5d", x[i][j]);
printf("\n");
}
}
void func(int x[][N], int n)
{
int i, j;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
if (i <= j)
x[i][j] = i + 1;
else
x[i][j] = j + 1;
}

任务六

task_1

#include<stdio.h>

#include<string.h>
#define N 80
int main() {
char views1[N] = "hey,c,I have not love u yet.";
char views2[N] = "hey,c,how can I love u?";
char t[N];
printf("交换前:\n");
printf("views1=%s\n", views1);
printf("views2=%s\n", views2);

strcpy_s(t, views1);
strcpy_s(views1, views2);
strcpy_s(views2, t);
printf("交换后:\n");
printf("views1:%s\n", views1);
printf("views2:%s\n", views2);
return 0;
}

task_2

#include<stdio.h>

#include<string.h>
#define N 80
int main() {
char views[2][N] = { "hey,c,I have not love u yet.","hey,c,how can I love u?" };
char t[N];
printf("交换前:\n");
printf("views1:%s\n", views[0]);
printf("views2:%s\n", views[1]);

strcpy_s(t, views[0]);
strcpy_s(views[0], views[1]);
strcpy_s(views[1], t);
printf("交换后:\n");
printf("views1:%s\n", views[0]);
printf("views2:%s\n", views[1]);
return 0;
}

任务七

#include<stdio.h>
#include<string.h>
#define N 5
#define M 20
void bubble_sort(char str[][M], int n);
int main()
{
char name[][M] = { "Bob","Bill","Joseph","Taylor","George" };
int i;
printf("输出初始名单:\n");
for (i = 0; i < N; i++)
printf("%s\n", name[i]);
printf("\n排序中...\n");
bubble_sort(name, N);
printf("\n按字典序输出名单:\n");
for (i = 0; i < N; i++)
printf("%s\n", name[i]);
return 0;
}

void bubble_sort(char str[][M], int n) {
int i, j;
char t[M];
for (j = 0; j < n - 1; j++)
for (i = 0; i < n - 1 - j; i++)
if (strcmp(str[i], str[i + 1]) > 0)
{
strcpy_s(t, str[i]);
strcpy_s(str[i], str[i + 1]);
strcpy_s(str[i + 1], t);
}
}