实验四

任务1.1

#include<stdio.h>
#define N 4

void test1(){
    int a[N]={1,9,8,4};
    int i;
    printf("sizeof(a)=%d\n",sizeof(a));
    for(i=0;i<N;i++)
        printf("%p:%d\n",&a[i],a[i]);
    printf("a = %p\n", a);
} 
void test2() {
    char b[N] = {'1', '9', '8', '4'};
    int i;
    printf("sizeof(b) = %d\n", sizeof(b));
    for (i = 0; i < N; ++i)
    printf("%p: %c\n", &b[i], b[i]);
    printf("b = %p\n", b);
} 
int main() {
    printf("测试1: int类型一维数组\n");
    test1();
    printf("\n测试2: char类型一维数组\n");
    test2();
    return 0;
}

结论:1)int型数组a在内存中是连续存放的，每个元素占用4个字节单位；不一样，前者返回的是数组的值，后者返回的是数组的地址。

　　 2)int型数组b在内存中是连续存放的，每个元素占用1个字节单位；不一样，前者返回的是数组的值，后者返回的是数组的地址。

任务1.2

#include <stdio.h>
#define N 2
#define M 4
void test1() {
    int a[N][M] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
    int i, j;

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

    printf("a = %p\n", a);
    printf("a[0] = %p\n", a[0]);
    printf("a[1] = %p\n", a[1]);
    printf("\n");
} 
void test2() {
    char b[N][M] = {{'1', '9', '8', '4'}, {'2', '0', '4', '9'}};
    int i, j;
    printf("sizeof(b) = %d\n", sizeof(b));
    for (i = 0; i < N; ++i)
        for (j = 0; j < M; ++j)
            printf("%p: %c\n", &b[i][j], b[i][j]);
    printf("\n");

    printf("b = %p\n", b);
    printf("b[0] = %p\n", b[0]);
    printf("b[1] = %p\n", b[1]);
} 
int main() {
    printf("测试1: int型两维数组");
    test1();
    printf("\n测试2: char型两维数组");
    test2();
    return 0;
}

结论:1)int型二维数组a，在内存中是按按行连续存放的，每个元素占用4个字节单元；不一样

　　 2)char型二维数组b，在内存中是按按行连续存放的，每个元素占用1个字节单元；不一样

　　 3)相差16；相差4；都是数组所占字节的一半

任务2

#include <stdio.h>
#include <string.h>
#define N 80
void swap_str(char s1[N], char s2[N]);
void test1();
void test2();
int main() {
printf("测试1: 用两个一维char数组，实现两个字符串交换\n");
test1();
printf("\n测试2: 用二维char数组，实现两个字符串交换\n");
test2();
return 0;
} 
void test1() {
char views1[N] = "hey, C, I hate u.";
char views2[N] = "hey, C, I love u.";
printf("交换前: \n");
puts(views1);
puts(views2);
swap_str(views1, views2);
printf("交换后: \n");
puts(views1);
puts(views2);
} 
void test2() {
char views[2][N] = {"hey, C, I hate u.",
"hey, C, I love u."};
printf("交换前: \n");
puts(views[0]);
puts(views[1]);
swap_str(views[0], views[1]);
printf("交换后: \n");
puts(views[0]);
puts(views[1]);
} 
void swap_str(char s1[N], char s2[N]) {
char tmp[N];
strcpy(tmp, s1);
strcpy(s1, s2);
strcpy(s2, tmp);
}

 

任务3.1

#include <stdio.h>
#define N 80
int count(char x[]);
int main() {
    char words[N+1];
    int n;
while(gets(words) != NULL) {
    n = count(words);
    printf("单词数: %d\n\n", n);
} 
    return 0;
} 
int count(char x[]) {
    int i;
    int word_flag = 0; 
    int number = 0; 
    for(i = 0; x[i] != '\0'; i++) {
    if(x[i] == ' ')
    word_flag = 0;
    else if(word_flag == 0) {
    word_flag = 1;
    number++;
}
} 
    return number;
}

任务3.2

#include <stdio.h>
#define N 1000
int main() {
    char line[N];
    int word_len;
    int max_len;
    int end;
    int i;
while(gets(line) != NULL) {
    word_len = 0;
    max_len = 0;
    end = 0;
    i = 0;
while(1) {

while(line[i] == ' ') {
    word_len = 0; 
    i++;
} 

while(line[i] != '\0' && line[i] != ' ') {
    word_len++;
    i++;
} 

if(max_len < word_len) {
    max_len = word_len;
    end = i;
} 

if(line[i] == '\0')
    break;
} 

    printf("最长单词: ");
for(i = end - max_len; i < end; ++i)
printf("%c", line[i]);
printf("\n\n");
}
    return 0;
}

任务4

#include <stdio.h>
#define N 100
void dec_to_n(int x, int n); 
int main() {
int x;
    printf("输入一个十进制整数: ");
while(scanf("%d", &x) != EOF) {
    dec_to_n(x, 2); 
    dec_to_n(x, 8); 
    dec_to_n(x, 16);
    printf("\n输入一个十进制整数: ");
} 
    return 0;
}
void dec_to_n(int x, int n){
    if(x==0){
        printf("0");
        return;
    }
    int a[100],cnt=0;
    while(x){
        a[cnt++]=x%n;
        x/=n;
    }
    for(int i=cnt-1;i>=0;i--){
        if(a[i]<10){
            printf("%d",a[i]);
        }
        else{
            printf("%c",a[i]-10+'A');
        }
    }printf("\n");
}

任务5

 

#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("%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 i;
    double sum=0.0;
    
    for(i=0;i<=n-1;i++){
        sum=sum+x[i]; 
    }
    
    return  sum/n;
} 
void bubble_sort(int x[], int n){
    int i,j,t;
    for(i=0;i<n-1;i++){
        for(j=0;j<n-1-i;j++){
            if(x[j]<x[j+1]){
                t=x[j];
                x[j]=x[j+1];
                x[j+1]=t;
            }
        }
    }
}

 

任务6

#include <stdio.h>
#include <string.h>
#define N 5
#define M 20

void output(char str[][M], int n);
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");
    output(name, N);
    printf("\n排序中...\n");
    bubble_sort(name, N); 
    printf("\n按字典序输出名单:\n");
    output(name, N);
    return 0;
}
void output(char str[][M], int n) {
    int i;
    for(i = 0; i < n; ++i)
        printf("%s\n", str[i]);
}
void bubble_sort(char str[][M], int n){
    int i,j;
    for(i=1;i<5;i++)   
 {
    for(j=0;j<5-i;j++)
    {
        if(strcmp(str[j],str[j+1])>0) 
        { 
             char str1[1000];
             strcpy(str1,str[j]); 
             strcpy(str[j],str[j+1]); 
             strcpy(str[j+1],str1); 
        }
     }
  }
}

任务7

 

#include <stdio.h>
#include<stdlib.h>
int main() {
    char a[101];
    int i,j,k=0;
    while(gets(a)!=NULL){
 
    for(i=0;i<101&&a[i]!='\0';++i)
    {  if(k==1) break;
        for(j=i+1;a[j]!='\0';++j)
     
        if(a[i]==a[j])
         
             k=1;
         
     
    }
    if(k==1) 
        printf("YES\n");
    else 
        printf("NO\n"); k=0;
    }
    system("pause");
    return 0;
}

 

任务8

 

#include <stdio.h>
#include<stdlib.h>
#include<string.h>
#define N 100
#define M 4
 

void output(int x[][N], int n);         
void rotate_to_right(int x[][N], int n);
 
 
int main() {
    int t[][N] = {{21, 12, 13, 24},
                  {25, 16, 47, 38},
                  {29, 11, 32, 54},
                  {42, 21, 33, 10}};
 
    printf("原始矩阵:\n");
    output(t, M); 
 
    rotate_to_right(t, M);
 
    printf("变换后矩阵:\n");
    output(t, M); 
    system("pause");
    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("%4d", x[i][j]);
 
        printf("\n");
    }
}
void rotate_to_right(int x[][N], int n){
         
        int i,t;
        for(i=0;i<4;++i){
        t=x[i][0];
        x[i][0]=x[i][3];
        x[i][3]=t;
        t=x[i][1];
        x[i][1]=x[i][3];
        x[i][3]=t;
        t=x[i][3];
        x[i][3]=x[i][2];
        x[i][2]=t;
        }
 
 
}