shiyan5

1_1

#include <stdio.h>
#define N 5

void input(int x[], int n);
void output(int x[], int n);
void find_min_max(int x[], int n, int *pmin, int *pmax);

int main() {
int a[N];
int min, max;

printf("录入%d个数据:\n", N);
input(a, N);

printf("数据是: \n");
output(a, N);

printf("数据处理...\n");
find_min_max(a, N, &min, &max);

printf("输出结果:\n");
printf("min = %d, max = %d\n", min, max);

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");
}

void find_min_max(int x[], int n, int *pmin, int *pmax) {
int i;

*pmin = *pmax = x[0];

for(i = 0; i < n; ++i)
if(x[i] < *pmin)
*pmin = x[i];
else if(x[i] > *pmax)
*pmax = x[i];
}

找最大最小值

指向x的位置

1_2

#include <stdio.h>
#define N 5

void input(int x[], int n);
void output(int x[], int n);
int *find_max(int x[], int n);

int main() {
int a[N];
int *pmax;

printf("录入%d个数据:\n", N);
input(a, N);

printf("数据是: \n");
output(a, N);

printf("数据处理...\n");
pmax = find_max(a, N);

printf("输出结果:\n");
printf("max = %d\n", *pmax);

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");
}

int *find_max(int x[], int n) {
int max_index = 0;
int i;

for(i = 0; i < n; ++i)
if(x[i] > x[max_index])
max_index = i;

return &x[max_index];
}

先找到最大数对应的序号，在更据序号找到最大值

可以

2_1

#include <stdio.h>
#include <string.h>
#define N 80

int main() {
char s1[N] = "Learning makes me happy";
char s2[N] = "Learning makes me sleepy";
char tmp[N];

printf("sizeof(s1) vs. strlen(s1): \n");
printf("sizeof(s1) = %d\n", sizeof(s1));
printf("strlen(s1) = %d\n", strlen(s1));

printf("\nbefore swap: \n");
printf("s1: %s\n", s1);
printf("s2: %s\n", s2);

printf("\nswapping...\n");
strcpy(tmp, s1);
strcpy(s1, s2);
strcpy(s2, tmp);

printf("\nafter swap: \n");
printf("s1: %s\n", s1);
printf("s2: %s\n", s2);

return 0;
}

23 sizeof算\0strlen不算

不，s1大小未申明，且字符串赋值要用strcpy

交换了

2_2

#include <stdio.h>
#include <string.h>
#define N 80

int main() {
char *s1 = "Learning makes me happy";
char *s2 = "Learning makes me sleepy";
char *tmp;

printf("sizeof(s1) vs. strlen(s1): \n");
printf("sizeof(s1) = %d\n", sizeof(s1));
printf("strlen(s1) = %d\n", strlen(s1));

printf("\nbefore swap: \n");
printf("s1: %s\n", s1);
printf("s2: %s\n", s2);

printf("\nswapping...\n");
tmp = s1;
s1 = s2;
s2 = tmp;

printf("\nafter swap: \n");
printf("s1: %s\n", s1);
printf("s2: %s\n", s2);

return 0;
}

存放字符串的初始地址，计算带\0和不带\0的字符串长度

不可，

交换的是指针，没

3

#include <stdio.h>

int main() {
int x[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
int i, j;
int *ptr1;
int(*ptr2)[4];

printf("输出1: 使用数组名、下标直接访问二维数组元素\n");
for (i = 0; i < 2; ++i) {
for (j = 0; j < 4; ++j)
printf("%d ", x[i][j]);
printf("\n");
}

printf("\n输出2: 使用指针变量ptr1(指向元素)间接访问\n");
for (ptr1 = &x[0][0], i = 0; ptr1 < &x[0][0] + 8; ++ptr1, ++i) {
printf("%d ", *ptr1);

if ((i + 1) % 4 == 0)
printf("\n");
}

printf("\n输出3: 使用指针变量ptr2(指向一维数组)间接访问\n");
for (ptr2 = x; ptr2 < x + 2; ++ptr2) {
for (j = 0; j < 4; ++j)
printf("%d ", *(*ptr2 + j));
printf("\n");
}

return 0;
}

先定义指针,在规定长度

定义的是指针数组

4

#include <stdio.h>
#define N 80

void replace(char *str, char old_char, char new_char); // 函数声明

int main() {
char text[N] = "Programming is difficult or not, it is a question.";

printf("原始文本: \n");
printf("%s\n", text);

replace(text, 'i', '*'); // 函数调用 注意字符形参写法，单引号不能少

printf("处理后文本: \n");
printf("%s\n", text);

return 0;
}

void replace(char *str, char old_char, char new_char) {
int i;

while(*str) {
if(*str == old_char)
*str = new_char;
str++;
}
}

如果不一样，用新的字符替换旧的

*str代表地址，所以不对

5

#include<stdio.h>
#include<string.h>
#define N 80
char *str_trunc(char *str,char x);
int main(){
char str[N];
char ch;
while(printf("输入字符串："),gets(str)!=NULL){
printf("输入一个字符：");
ch=getchar();
printf("截断处理...\n");
str_trunc(str,ch);
printf("截断处理后的字符串:%S\n\n",str);
getchar();
}
return 0;
}
char *str_trunc(char *str,char x){
int i=0;
char a[N];
do{
a[i]=str[i];
i++;
}while(str[i]!=x);
a[i]='\0';
strcpy(str,a);
return str;
}

6

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

int check_id(char *str);

int main()
{
char *pid[N] = {"31010120000721656X",
"3301061996X0203301",
"53010220051126571",
"510104199211197977",
"53010220051126133Y"};
int i;

for (i = 0; i < N; ++i)
if (check_id(pid[i]))
printf("%s\tTrue\n", pid[i]);
else
printf("%s\tFalse\n", pid[i]);

return 0;
}

int check_id(char *str) {
int len=strlen(str),i;
if(len!=18){
return 0;
}
for(i=0;i<len;i++)
while(i<17){
if(str[i]>'9')
return 0;
}
while(i=17){
if(str[i]!='X')
return 0;}
return 1;

}

7

#include <stdio.h>
#include<string.h>
#define N 80
void encoder(char *str, int n);
void decoder(char *str, int n);

int main() {
char words[N];
int n;

printf("输入英文文本: ");
gets(words);

printf("输入n: ");
scanf("%d", &n);

printf("编码后的英文文本: ");
encoder(words, n);
printf("%s\n", words);

printf("对编码后的英文文本解码: ");
decoder(words, n);
printf("%s\n", words);

return 0;
}

void encoder(char *str, int n) {
int i,len;
len=strlen(str);
for(i=0;i<len;i++)
if(str[i]>='a'&&str[i]<='z')
str[i]=((str[i]-'a'+n)%26)+'a';
if(str[i]>='A'&&str[i]<='Z')
str[i]=((str[i]-'A'+n)%26)+'A';
}

void decoder(char *str, int n) {
int i,len;
len=strlen(str);
for(i=0;i<len;i++)
if(str[i]>='a'&&str[i]<='z')
str[i]=((str[i]-'a'-n+26)%26)+'a';
if(str[i]>='A'&&str[i]<='Z')
str[i]=((str[i]-'a'-n+26)%26)+'A';
}

8

把上课讲的方法忘了，对不起