实验四的详细步骤是怎样的？
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本文共计1120个文字，预计阅读时间需要5分钟。
Task 1-1: 简化创新以下开头内容，不超过100字。
创新是推动社会进步的核心动力，它源自于人类对未知领域的探索和突破。在科技日新月异的今天，创新已成为国家竞争的关键要素。
task 1-1#include <stdio.h>
#define N 4
int main()
{
int a[N] = {2, 0, 2, 2};
char b[N] = {'2', '0', '2', '2'};
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;
}
①int型数组a在内存中连续存放，每个元素占用4个内存字节单元
②char型数组b在内存中连续存放，每个元素占用一个内存字节单元。
③数组名a对应的值和&a[0]一样，数组b对应的值和&b[0]是一样的
task 1-2#include <stdio.h>
#define N 2
#define M 3
int main()
{
int a[N][M] = {{1, 2, 3}, {4, 5, 6}};
char b[N][M] = {{'1', '2', '3'}, {'4', '5', '6'}};
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按行连续存放，每个元素占4个内存字节单位
②按行连续存放，每个元素占用1个内存字节单元。
task 2#include<stdio.h>
int days_of_year(int year, int month, int day);
int main()
{
int year, month, day;
int days;
while(scanf("%d%d%d", &year, &month, &day) != 0)
{
days = days_of_year(year, month, day);
printf("%4d-%02d-%02d是这一年中的第%d天.\n\n", year, month, day, days);
}
return 0;
}
int days_of_year(int year, int month, int day)
{
int n=0,i;
int x[13]={0,31,28,31,30,31,30,31,31,30,31,30,31 };
if ((year%4==0&&year%100!=0)||(year%400==0))
x[2]=29;
for (i=1;i<month;i++)
n=n+x[i];
return n+day;
}
#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 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输入课程分数：\n");
output(scores, N);
printf("\n课程分数处理：计算均分、排序...\n");
ave = average(scores, N);
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 sum = 0, i;
double ave;
for (i = 0; i < n; ++i)
{
sum += x[i];
}
ave = 1.0*sum / N;
return ave;
}
void sort(int x[], int n)
{
int i, j, temp;
for (i = 1; i < n; i++)
{
for (j = 0; j < n-1; j++)
{
if (x[j] < x[j + 1])
{
temp = x[j];
x[j] = x[j + 1];
x[j + 1] = temp;
}
}
}
}
#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)
{
if (x == 0)
printf("\n");
else
{
dec2n(x / n, n);
if (n >= 10)
{
if (x % n >= 10)
printf("%c\n", (char)((x % n - 10) + 'A'));
else
printf("%d", x % n);
}
else
printf("%d", x % n);
}
}
#include<stdio.h>
#define N 100
int main()
{
int n, i, j, x[N][N];
printf("Enter n：");
while (scanf("%d", &n) != 0)
{
for (i = 0; i <= n - 1; i++)
{
for (j = 0; j <= n - 1; j++)
{
if (i <= j)
x[i][j] = i + 1;
else
x[i][j] = j + 1;
printf("%2d", x[i][j]);
}
printf("\n");
}
printf("\nEnter n：");
}
return 0;
}
#include<stdio.h>
void output(char x[]);
void sort(char x[], char y[]);
#define N 80
int main()
{
char views1[N] = "hey,c,i hate u.";
char views2[N] = "hey,c,i love u.";
printf("original views:");
printf("\nviews1：");
output(views1);
printf("\nviews2：");
output(views2);
printf("\n\nswapping...\n");
sort(views1, views2);
printf("\nviews1：");
output(views1);
printf("\nviews2：");
output(views2);
return 0;
}
void output(char x[])
{
for (int i=0; x[i] != '\0'; i++)
printf("%c", x[i]);
}
void sort(char x[], char y[])
{
int temp;
for (int i = 0; x[i] != '\0'; i++)
{
temp = x[i];
x[i] = y[i];
y[i] = temp;
}
}
#define _CRT_SECURE_NO_WARNINGS
#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)
{
char temp[M];
for (int i = 0; i < n - 1; i++)
{
for (int j = 0; j < n - 1; j++)
{
if (strcmp(str[j], str[j + 1]) > 0)
{
strcpy(temp, str[j]);
strcpy(str[j], str[j + 1]);
strcpy(str[j + 1], temp);
}
}
}
}
本文共计1120个文字，预计阅读时间需要5分钟。
Task 1-1: 简化创新以下开头内容，不超过100字。
创新是推动社会进步的核心动力，它源自于人类对未知领域的探索和突破。在科技日新月异的今天，创新已成为国家竞争的关键要素。
task 1-1#include <stdio.h>
#define N 4
int main()
{
int a[N] = {2, 0, 2, 2};
char b[N] = {'2', '0', '2', '2'};
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;
}
①int型数组a在内存中连续存放，每个元素占用4个内存字节单元
②char型数组b在内存中连续存放，每个元素占用一个内存字节单元。
③数组名a对应的值和&a[0]一样，数组b对应的值和&b[0]是一样的
task 1-2#include <stdio.h>
#define N 2
#define M 3
int main()
{
int a[N][M] = {{1, 2, 3}, {4, 5, 6}};
char b[N][M] = {{'1', '2', '3'}, {'4', '5', '6'}};
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按行连续存放，每个元素占4个内存字节单位
②按行连续存放，每个元素占用1个内存字节单元。
task 2#include<stdio.h>
int days_of_year(int year, int month, int day);
int main()
{
int year, month, day;
int days;
while(scanf("%d%d%d", &year, &month, &day) != 0)
{
days = days_of_year(year, month, day);
printf("%4d-%02d-%02d是这一年中的第%d天.\n\n", year, month, day, days);
}
return 0;
}
int days_of_year(int year, int month, int day)
{
int n=0,i;
int x[13]={0,31,28,31,30,31,30,31,31,30,31,30,31 };
if ((year%4==0&&year%100!=0)||(year%400==0))
x[2]=29;
for (i=1;i<month;i++)
n=n+x[i];
return n+day;
}
#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 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输入课程分数：\n");
output(scores, N);
printf("\n课程分数处理：计算均分、排序...\n");
ave = average(scores, N);
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 sum = 0, i;
double ave;
for (i = 0; i < n; ++i)
{
sum += x[i];
}
ave = 1.0*sum / N;
return ave;
}
void sort(int x[], int n)
{
int i, j, temp;
for (i = 1; i < n; i++)
{
for (j = 0; j < n-1; j++)
{
if (x[j] < x[j + 1])
{
temp = x[j];
x[j] = x[j + 1];
x[j + 1] = temp;
}
}
}
}
#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)
{
if (x == 0)
printf("\n");
else
{
dec2n(x / n, n);
if (n >= 10)
{
if (x % n >= 10)
printf("%c\n", (char)((x % n - 10) + 'A'));
else
printf("%d", x % n);
}
else
printf("%d", x % n);
}
}
#include<stdio.h>
#define N 100
int main()
{
int n, i, j, x[N][N];
printf("Enter n：");
while (scanf("%d", &n) != 0)
{
for (i = 0; i <= n - 1; i++)
{
for (j = 0; j <= n - 1; j++)
{
if (i <= j)
x[i][j] = i + 1;
else
x[i][j] = j + 1;
printf("%2d", x[i][j]);
}
printf("\n");
}
printf("\nEnter n：");
}
return 0;
}
#include<stdio.h>
void output(char x[]);
void sort(char x[], char y[]);
#define N 80
int main()
{
char views1[N] = "hey,c,i hate u.";
char views2[N] = "hey,c,i love u.";
printf("original views:");
printf("\nviews1：");
output(views1);
printf("\nviews2：");
output(views2);
printf("\n\nswapping...\n");
sort(views1, views2);
printf("\nviews1：");
output(views1);
printf("\nviews2：");
output(views2);
return 0;
}
void output(char x[])
{
for (int i=0; x[i] != '\0'; i++)
printf("%c", x[i]);
}
void sort(char x[], char y[])
{
int temp;
for (int i = 0; x[i] != '\0'; i++)
{
temp = x[i];
x[i] = y[i];
y[i] = temp;
}
}
#define _CRT_SECURE_NO_WARNINGS
#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)
{
char temp[M];
for (int i = 0; i < n - 1; i++)
{
for (int j = 0; j < n - 1; j++)
{
if (strcmp(str[j], str[j + 1]) > 0)
{
strcpy(temp, str[j]);
strcpy(str[j], str[j + 1]);
strcpy(str[j + 1], temp);
}
}
}
}