curlen函数C语言,curl c

#include #define MaxLen 100 typedef struct { char str[MaxLen]; int curlen; }string; string

#include stdio.h

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#define MaxSize 100

typedef struct

{

char data[MaxSize];

int len; //串长

} SqString;

void StrAssign(SqString str,char cstr[]) //str为引用型参数

{

int i;

for (i=0;cstr[i]!='\0';i++)

str.data[i]=cstr[i];

str.len=i;

}

void StrCopy(SqString s,SqString t) //s为引用型参数

{

int i;

for (i=0;it.len;i++)

s.data[i]=t.data[i];

s.len=t.len;

}

int StrLength(SqString s)

{

return s.len;

}

SqString Concat(SqString s,SqString t)

{

SqString str;

int i;

str.len=s.len+t.len;

for (i=0;is.len;i++) //将s.data[0]～s.data[s.len-1]复制到str

str.data[i]=s.data[i];

for (i=0;it.len;i++) //将t.data[0]～t.data[t.len-1]复制到str

str.data[s.len+i]=t.data[i];

return str;

}

SqString SubStr(SqString s,int i,int j)

{

SqString str;

int k;

str.len=0;

if (i=0 || is.len || j0 || i+j-1s.len)

return str; //参数不正确时返回空串

for (k=i-1;ki+j-1;k++) //将s.data[i]～s.data[i+j]复制到str

str.data[k-i+1]=s.data[k];

str.len=j;

return str;

}

void DispStr(SqString s)

{

int i;

if (s.len0)

{

for (i=0;is.len;i++)

printf("%c",s.data[i]);

printf("\n");

}

}

SqString invert(SqString s) //例6-2的算法

{

SqString s1,s2;

if (StrLength(s)0)

{

s1=invert(SubStr(s,2,StrLength(s)-1));

s2=Concat(s1,SubStr(s,1,1));

}

else

StrCopy(s2,s);

return s2;

}

void main()

{

SqString s,t;

StrAssign(s,"abcd1234");

printf("s:");DispStr(s);

printf("s-t\n");

t=invert(s);

printf("t:");DispStr(t);

顺序查找线性表的主函数怎么写?

假设前面定义过链表p了，待查找的是n。

q=p;

while (q-next!=null *q!=n)

q=q-next;

if (*q!=n) printf("no suitable value")

求数据结构（c语言版）程序源代码

1 #include string.h

2 #include stdio.h

3 #include stdlib.h

4

5 #define MAX_POS_NUM 100

6 #define MAX_STR_LEN 1024

7

8

9 //1. get all position of str_z in str_x

10 int get_sub_str_pos(const char * str_x, const char * str_z, int sub_str_pos[])

11 {

12 if (NULL == str_x || NULL == str_z)

13 {

14 printf("in error!\n");

15 return -1;

16 }

17

18 const char * pos_ptr = NULL;

19

20 pos_ptr = strstr(str_x,str_z);

21

22 int i=0;

23 while(pos_ptr)

24 {

25 printf("substring positon:%d\n",pos_ptr-str_x+1);

26 sub_str_pos[i] = pos_ptr - str_x + 1;

27 pos_ptr = strstr(pos_ptr+strlen(str_z),str_z);

28 i++;

29 }

30

31 return 0;

32 }

33

34 //2. get max length common string of str_x and str_y

35 char * get_max_com_str(const char * str_x, const char * str_y)

36 {

37 int x_len = strlen(str_x);

38 int y_len = strlen(str_y);

39

40 char * tmp_str = new char[y_len+1];

41

42 for(int i=y_len; i0; i--) // i is substring length

43 {

44 if (ix_len)

45 continue;

46 for(int j=0;j=y_len-i; j++) // j is substring start postion

47 {

48 snprintf(tmp_str,i+1,"%s",str_y);

49 if (strstr(str_x,tmp_str))

50 {

51 printf("%s\n",tmp_str);

52 printf("max common substring length:%d\n",i);

53 return tmp_str;

54 }

55 }

56 }

57

58 return NULL;

59 }

60

61 //3. replace all substring in question 1

62 char * replace_sub_str(const char * str_x, char * max_com_str, int sub_str_pos[], int sub_str_len)

63 {

64 char * replaced_str = new char[MAX_STR_LEN];

65

66 int sub_pos = sub_str_pos[0];

67 int l=0; // l is sub_str_pos index

68 int i=0,j=0; //i is str_x pos, j is replaced_str pos

69

70 while(*str_x)

71 {

72 if (i==sub_pos-1) // replace from this position

73 {

74 // printf ("i:%d,\n",i);

75 for (int k=0; kstrlen(max_com_str); k++)

76 {

77 *(replaced_str + j) = * (max_com_str + k);

78 j++;

79 }

80 i += sub_str_len;

81 str_x += sub_str_len;

82 l++;

83 sub_pos = sub_str_pos[l];

84 continue;

85 }

86 *(replaced_str+j) = *str_x++;

87 i++;

88 j++;

89 }

90

91 * (replaced_str + j) = '\0';

92

93 return replaced_str;

94 }

95

96 int main()

97 {

98 const char * str_x = "abcabcabc";

99 const char * str_y = "cabcd";

100 const char * str_z = "abc";

101

102 int sub_str_pos [MAX_POS_NUM] = {0};

103

104 char * max_com_str = NULL;

105

106 char * replaced_str = NULL;

107

108 get_sub_str_pos(str_x,str_z,sub_str_pos);

109 max_com_str = get_max_com_str(str_x,str_y);

110

111 printf("max common str: %s\n",max_com_str);

112

113 replaced_str = replace_sub_str(str_x, max_com_str, sub_str_pos, strlen(str_z));

114 printf("repalced str: %s\n",replaced_str);

115

116 return 0;

117 }

C++设计：一元多项式相加

#includeiostream.h

#includemalloc.h

#define len sizeof(LNode)

typedef struct Lnode //结点结构声明

{ int nf; // 多项式系数

int ne; //多项式幂指数

struct LNode *next;

}LNode;

typedef LNode *Pol;

LNode *creat(void) //创建多项式函数

{ LNode *head,*q;

LNode *n;

head=q=n=(LNode *)malloc(len);

q-next=NULL;

do{n=(LNode *)malloc(len);

cinn-nen-nf;

q-next=n;

q=n;}while(n-nf!=0||n-ne!=0);

q-next=NULL;

return head;}

LNode *output(LNode *head) //输出多项式的函数

{LNode *p;

p=head-next;int n=0;

if(p-nf==0p-ne==0) {cout"0";return 0;}

else do{if(p-nf!=0)

{if(n==0)

{if(p-ne==0) {coutp-nf;break;}

else{coutp-nf"X"p-ne;n++;p=p-next;}}

else {if(p-ne==0) {if(p-nf0)coutp-nf;else cout"+"p-nf;break;}

else if(p-nf0) {coutp-nf"X"p-ne;p=p-next;}

else {cout"+"p-nf"X"p-ne;p=p-next;}}}

else p=p-next;}while(p-nf!=0||p-ne!=0);

if(n==0p-nf==0p-ne==0) {cout"0";return 0;}

return 0;}

LNode * add(Pol Pa,Pol Pb)//两个多项式相加的函数

{LNode *p1,*p2,*p,*pr,*p0;

p1=Pa-next;p2=Pb-next;

p0=pr=Pa;

while(p1-next!=NULL||p2-next!=NULL)

{if(p1-nep2-ne)

{pr=p1;p1=p1-next;}

else if(p1-ne==p2-ne)

{p1-nf=p1-nf+p2-nf;p2=p2-next;}

else {pr-next=p2;p2=p2-next; pr=pr-next;pr-next=p1;}

if(p1-ne==0p2-next==NULL) break;}

p1-next=p2;

return p0;

}

void main()//主函数

{char y;

for(;;){Pol p1,p2,p3;

cout"input ha"endl;

p1=creat(); //建立多项式HA

cout"input hb"endl;

p2=creat(); //建立多项式HB

cout"ha=";*output(p1);coutendl;//输出多项式HA

cout"hb=";*output(p2);coutendl; //输出多项式HB

p3=add(p1,p2); //多项式HA与HB相加

cout"ha+hb=";*output(p3);coutendl; 输出多项式HC

cout"ARE YOU CONTINUE?(Y|N)"endl;

ciny;

if(y=='n') break;}

写一函数，输入一行字符，将次字符串的最长单词输出---C语言

#include stdio.h

#include string.h

int main()

{

char szText[100] = {0};

int nMaxBegin = 0, nMaxLen = 0, nCurLen = 0;

gets(szText);

szText[strlen(szText)] = ' ';

for (int i = 0; i strlen(szText); i++)

{

if (szText[i] != ' ')

{

nCurLen ++;

}

else

{

if (nMaxLen nCurLen)

{

nMaxBegin = i - nCurLen;

nMaxLen = nCurLen;

}

nCurLen = 0;

}

}

char szMaxString[100] = {0};

memcpy(szMaxString, szText + nMaxBegin, nMaxLen);

printf("%s\n", szMaxString);

return 0;

}

C语言字符串如何循环输入与储存

#includestdio.h

#includestring.h

char str[1005],ans[1005];

void main()

{

int MaxLen = 0;

int CurLen;

int i;

for(i=0;i10;++i)

{

gets(str);

CurLen = strlen(str);

if(CurLenMaxLen)MaxLen=CurLen,strcpy(ans,str);

}

puts(ans);

}

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