c语言strcm二维数组用法,c08，c09二维数组，自己的字符串函数，strcpy，strcat，strncat，strncpy，atoi，itoa，strcmp...

二维数组，魔方阵，杨辉三角

1.打印函数

void Show(int (*brr)[3],int row,int col)

{

for(int i = 0;i < row;i++)

{

for(int j = 0;j < col;j++)

{

printf("%d ",brr[i][j]);

}

printf("\n");

}

}

void Show2(int *brr,int row,int col)//方法二

{

for(int i = 0;i < row;i++)//1

{

for(int j = 0;j < col;j++)//3

{

printf("%2d ",brr[i*col+j]);//1*3+0

}

printf("\n");

}

}

int main()

{

int brr[ROW][ROW] = {};

Fun2(brr);

Show2((int *)brr,ROW,ROW);

}

2.对称矩阵

//对称矩阵

void Fun2(int (*brr)[ROW])

{

srand(time(0));//随机种子

for(int i = 0;i < ROW;i++)

{

for(int j = 0;j < ROW;j++)

{

if(j > i)

{

brr[i][j] = rand()%10+1;//[0-n)

}

else if(i == j)

{

brr[i][j] = 0;

}

else

{

brr[i][j] = brr[j][i] ;

}

}

}

}

注意

//int (*brr)[3] 数组指针

//int *brr[3] 指针数组

3.注意

int main()

{

int brr[2][3] = {1,2,3,4,5,6};

printf("brr == %d\n",brr);

printf("brr+1 ==%d\n",brr+1);

printf("*brr+1 == %d\n",*brr+1);

printf("*(brr+1) == %d\n",*(brr+1));

printf("*(brr+1)+2===>%d\n",*(brr+1)+2);

printf("&brr[1][2]==>%d\n",&brr[1][2]);

printf("*(*(brr+1)+2)====%d\n",*(*(brr+1)+2));

printf("*(brr[1]+2)===>%d\n",*(brr[1]+2));

return 0；

}

4. 练习题(杨辉三角，魔方阵)

#include

#include

#include

#define ROW 3

void Show2(int *brr,int row,int col)

{

for(int i=0;i

{

for(int j=0;j<=i;j++)

{

printf("%2d",brr[i*row+j]);

}

printf("\n");

}

}

void Show(int *brr,int row,int col)

{

for (int i = 0; i < ROW; i++)

{

for (int j = 0; j < ROW; j++)

{

printf("%3d",brr[i*row+j]);

}

printf("\n"); //打印一行回车

}

}

//作业杨辉三角

void YangHuiTriangle(int (*brr)[ROW])//数组指针

{

brr[0][0]=1;

for(int i=1;i

{

for(int j=0;j

{

if (j==0)//第0列为1

{

brr[i][j]=1;

}

else if(i==j)//对角线为1

{

brr[i][j]=1;//brr=brr+i;

}

else

{

brr[i][j]=brr[i-1][j-1]+brr[i-1][j];//杨辉三角的公式

}

}

}

}

魔方阵

//魔方阵奇数行列对角线加起来相等，1放在第0行中间位置。每次放在上一行下一列，如果当前数字的上一行下一列列有数据，则放在数字的下一行

void MagicMquare1(int (*brr)[ROW])

{

assert(ROW%2!=0);

int count =1;

int row=0;

int col=ROW/2;

while(count<=ROW*ROW)

{

brr[row][col]=count;

int i=row;

int j=col;

if(i==0)//如果i为0，则放到ROW-1行

{

i=ROW-1;

}

else//i!=0，放到i-1行

{

i=i-1;

//j=j+1;

}

j=(j+1)%ROW;//保证j是有效行,即j为可以+1的行

if(brr[i][j]!=0||(row==0&&col==ROW-1))//如果有值，放在当前的下一行

{

i=row+1;

j=col;

}

row=i;

col=j;

count ++;

}

/*for (int i = 0; i < ROW; i++)

{

for (int j = 0; j < ROW; j++)

{

printf("%3d",brr[i][j]);

}

printf("\n");

}*/

}

int main()

{

int brr[ROW][ROW]={0};

//YangHuiTriangle(brr);

//Show2((int *)brr,5,5);

MagicMquare1(brr);

Show((int *)brr,3,3);

}

void MagicSqaure(int (*arr)[ROW])//法二

{

assert(ROW % 2 != 0);

arr[0][ROW/2] = 1;

int preRow = 0;

int preCol = ROW/2;

for(int i = 2;i <= ROW*ROW;i++)

{

if( arr[(preRow-1+ROW) %ROW] [(preCol+1)%ROW] != 0 )

{

preRow = (preRow+1) % ROW ;

}

else

{

preRow = (preRow-1+ROW) %ROW;

preCol = (preCol+1)%ROW;

}

arr[preRow][preCol] = i;

}

}

5.My_strcpy，strcat，strncat，strncpy，atoi，itoa，strcmp

5.1My_strcpy，My strncpy

char * My_strcpy4(char *dest,const char *src)

{

//char *p = dest;

assert(dest != NULL && src != NULL);

while(*dest++ = *src++) {}

return dest;//走到最后\0

//return p;

}

#include

#include

#include

char *My_strncpy( char *dest,char *src,int n)

{

char *p=dest;

int count=0;

assert(dest!=NULL&&src!=NULL);

assert(n>0&&n

while(count

{

*dest++=*src++;

count++;

}

return p;

}

int main()

{

char dest[10]="";

printf("%s\n",My_strncpy(dest,"abcdefgh",5));

return 0;

}

5.2My strcat，My strncat

char *My_strcat(char *dest,char *src,int len)

{

assert(dest!=NULL&&src!=NULL);

if(strlen(dest)+strlen(src)>len);

char *p=dest;

while(*dest!='\0')

{

dest++;

}

while(*dest++=*src++);

return p;

}

int main()

{

char dest[10]="abc";

char *src="def";

int len=sizeof(dest)/sizeof(dest)[0];

printf("%s\n",My_strcat(dest,src,len));

return 0;

}

5.3My strcmp

int Mystrcmp(const char *str1,const char *str2)

{

assert(str1!=NULL&&str2!=NULL);

int tmp=0;

while((tmp=*str1-*str2)==0&&*str2!='\0')

{

str1++;

str2++;

}

return tmp;

}

int main()

{

char *str1="adefs";

char *str2="aadek";

printf("%d\n",Mystrcmp(str1,str2));

return 0;

}

5.4 atoi，itoa

#include

#include

#include

#include

int My_atoi(char *str)

{

int p=0;

assert(str!=NULL);

while(*str==' '||!isdigit(*str))

{

str++;

}

while(*str=='-')

{

printf("%s\n",'-');

}

while(isdigit(*str))

{

p=p*10+(*str-'0');

str++;

}

return p;

}

int main()

{

printf("%d\n",My_atoi(" daa-123"));

return 0;

}

char *My_itoa(char *str,int num)

{

int i=0;

if(num<0)

{

num=-num;

}

while(num!=0)

{

str[i]=num%10+'0';

i++;

num/=10;

}

str[i]='\0';

i--;

for(int j=0;j

{

char ch=str[j];

str[j]=str[i];

str[i]=ch;

}

return str;

}

int main()

{

char str[10]=" ";

printf("%s\n",My_itoa(str,123));

return 0;

}

5.5找出和为给定数值的两个元素的下标

void Sumk(int *arr,int key,int *num1,int *num2,int len)

{

int low = 0;

int high = len-1;

while(low < high)

{

if(arr[low]+arr[high] > key)

{

high--;

}

else if(arr[low]+arr[high] < key)

{

low++;

}

else

{

*num1 = low;

*num2 = high;

break;

}

}

}

int main()

{

int arr[]={4,6,8,9,10,11,13,23};

int len=sizeof(arr)/sizeof(arr[0]);

int index1 = 0;

int index2 = 0;

Sumk(arr,20,&index1,&index2,len);

printf("%d,%d\n",index1,index2);

//Print();

//printf("%d\n",CheckSum());

return 0;

}

5.6统计单词的个数

#include

#include

#include

int NumWord(char *str)

{

assert(str!=NULL);

int count=0;

while(*str!='\0')

{

if(isalpha(*str)&&!isalpha(*str+1))

{

count++;

}

str++;

}

}

int main()

{

printf("%\c",NumWord("dgs jsb jkd sjd sk jd k"));

}