本文目录一览：
- 1、c语言编程，请高手帮忙！有最高加分！！！200
- 2、C语言的经典编程例子
- 3、求C语言程序设计实例（200行）
- 4、c语言编程:找出200以内所有完数,并输出其因子 输入输出示例：1=2 6=1+2+3 28=1+2+4+7+14
- 5、c语言编程代码
c语言编程，请高手帮忙！有最高加分！！！200
我做到了，希望楼主也加到满分哟，呵呵。
程序在找子串时，如在abcccc中找子串cc时结果是3个而不是2个，当然楼主如果要求是2个稍改下就成。
输入字符串时允许空格。
用法：
1 输入一串字符，如123 abvc uuuu(回车）
2 程序显示统计有多少字符的结果，并要求输入想查的子串此时输入想查的子串，如 uu（回车）
3 程序显示有3个uu子串在里边。
注：我规定的是只有字母才算字符。
myfile.txt生成在D盘上。
编译环境TC2.0
#include"stdio.h"
#include"stdlib.h"
void main()
{
FILE *fp;
char ch,*p,*pcopy,*pp,a[1000],*tiaojian;
int i=0;
if((fp=fopen("d:\\myfile.txt","w+"))==NULL)
{
printf("Cannot open file strike any key exit!");
getch();
exit(1);
}
printf("please input the sentence here : \n");
gets(p);
while(*p!='\0') {
ch=*p++;
fputc(ch,fp);} /*写到myfile.txt*/
fclose(fp);
if((fp=fopen("d:\\myfile.txt","r+"))==NULL) /*从myfile.txt中读取文件*/
{
printf("Cannot open file strike any key exit!");
getch();
exit(1);
}
p=a;
while ((*p++=fgetc(fp))!=EOF) {
if ((*(p-1)='z' *(p-1)='a') || (*(p-1)='Z' *(p-1)='A')) i++;}
*p='\0';
printf("\n\nCOUNT:there are %d words in your sentence!",i);
p=a;
printf("\n\ninput the words that you want to look for: \n");
fflush(stdin);
gets(tiaojian);
pp=tiaojian;
i=0;
while (*p!='\0') {
pcopy=p++;
while (*pcopy++==*tiaojian++) {if (*tiaojian=='\0') {i++;break;}}
tiaojian=pp; }
printf("\n\n%d ge zi chuan!",i);
fclose(fp);
getch();
}
C语言的经典编程例子
//最经典的当然是HelloWorld了。
#include "stdio.h"
int main(void)
{
printf("HelloWorld!\r\n");
}
求C语言程序设计实例（200行）
时间函数举例程序分析
2.程序源代码：
#include "stdio.h"
#include "time.h"
void main()
{
time_t lt; /*define a longint time varible*/
lt=time(NULL);/*system time and date*/
printf(ctime()); /*english format output*/
printf(asctime(localtime()));/*tranfer to tm*/
printf(asctime(gmtime())); /*tranfer to Greenwich time*/
}
【程序92】
题目：时间函数举例2
1.程序分析：
2.程序源代码：
/*calculate time*/
#include "time.h"
#include "stdio.h"
main()
{
time_t start,end;
int i;
start=time(NULL);
for(i=0;i3000;i++)
{
printf("\1\1\1\1\1\1\1\1\1\1\n");
}
end=time(NULL);
printf("\1: The different is %6.3f\n",difftime(end,start));
}
【程序93】
题目：时间函数举例3
1.程序分析：
2.程序源代码：
/*calculate time*/
#include "time.h"
#include "stdio.h"
main()
{
clock_t start,end;
int i;
double var;
start=clock();
for(i=0;i10000;i++)
{
printf("\1\1\1\1\1\1\1\1\1\1\n");
}
end=clock();
printf("\1: The different is %6.3f\n",(double)(end-start));
}
【程序94】
题目：时间函数举例4,一个猜数游戏，判断一个人反应快慢。（版主初学时编的）
1.程序分析：
2.程序源代码：
#include "time.h"
#include "stdlib.h"
#include "stdio.h"
main()
{
char c;
clock_t start,end;
time_t a,b;
double var;
int i,guess;
srand(time(NULL));
printf("do you want to play it.('y' or 'n') \n");
loop:
while((c=getchar())=='y')
{
i=rand()%100;
printf("\nplease input number you guess:\n");
start=clock();
a=time(NULL);
scanf("%d",guess);
while(guess!=i)
{
if(guessi)
{
printf("please input a little smaller.\n");
scanf("%d",guess);
}
else
{
printf("please input a little bigger.\n");
scanf("%d",guess);
}
}
end=clock();
b=time(NULL);
printf("\1: It took you %6.3f seconds\n",var=(double)(end-start)/18.2);
printf("\1: it took you %6.3f seconds\n\n",difftime(b,a));
if(var15)
printf("\1\1 You are very clever! \1\1\n\n");
else if(var25)
printf("\1\1 you are normal! \1\1\n\n");
else
printf("\1\1 you are stupid! \1\1\n\n");
printf("\1\1 Congradulations \1\1\n\n");
printf("The number you guess is %d",i);
}
printf("\ndo you want to try it again?(\"yy\".or.\"n\")\n");
if((c=getch())=='y')
goto loop;
}
【程序95】
题目：家庭财务管理小程序
1.程序分析：
2.程序源代码：
/*money management system*/
#include "stdio.h"
#include "dos.h"
main()
{
FILE *fp;
struct date d;
float sum,chm=0.0;
int len,i,j=0;
int c;
char ch[4]="",ch1[16]="",chtime[12]="",chshop[16],chmoney[8];
pp: clrscr();
sum=0.0;
gotoxy(1,1);printf("|----------------------------------------------------|");
gotoxy(1,2);printf("| money management system(C1.0) 2000.03 |");
gotoxy(1,3);printf("|----------------------------------------------------|");
gotoxy(1,4);printf("| -- money records -- | -- today cost list -- |");
gotoxy(1,5);printf("| ------------------------ |-----------------------------|");
gotoxy(1,6);printf("| date: -------------- | |");
gotoxy(1,7);printf("| | | | |");
gotoxy(1,8);printf("| -------------- | |");
gotoxy(1,9);printf("| thgs: ------------------ | |");
gotoxy(1,10);printf("| | | | |");
gotoxy(1,11);printf("| ------------------ | |");
gotoxy(1,12);printf("| cost: ---------- | |");
gotoxy(1,13);printf("| | | | |");
gotoxy(1,14);printf("| ---------- | |");
gotoxy(1,15);printf("| | |");
gotoxy(1,16);printf("| | |");
gotoxy(1,17);printf("| | |");
gotoxy(1,18);printf("| | |");
gotoxy(1,19);printf("| | |");
gotoxy(1,20);printf("| | |");
gotoxy(1,21);printf("| | |");
gotoxy(1,22);printf("| | |");
gotoxy(1,23);printf("|--------------------------------------------------|");
i=0;
getdate(d);
sprintf(chtime,"%4d.%02d.%02d",d.da_year,d.da_mon,d.da_day);
for(;;)
{
gotoxy(3,24);printf(" Tab __browse cost list Esc __quit");
gotoxy(13,10);printf(" ");
gotoxy(13,13);printf(" ");
gotoxy(13,7);printf("%s",chtime);
j=18;
ch[0 ]=getch();
if(ch[0]==27)
break;
strcpy (chshop,"");
strcpy(chmoney,"");
if(ch[0]==9)
{
mm:i=0;
fp=fopen("home.dat","r+");
gotoxy(3,24);printf(" ");
gotoxy(6,4);printf(" list records ");
gotoxy(1,5);printf("|-------------------------------------|");
gotoxy(41,4);printf(" ");
gotoxy(41,5);printf(" |");
while(fscanf(fp,"%10s%14s%f\n",chtime,chshop,chm)!=EOF)
{
if(i==36)
{
getch();
i=0;
}
if ((i%36)17)
{
gotoxy(4,6+i);
printf(" ");
gotoxy(4,6+i);
}
else
if((i%36)16)
{
gotoxy(41,4+i-17);
printf(" ");
gotoxy(42,4+i-17);
}
i++;
sum=sum+chm;
printf("%10s %-14s %6.1f\n",chtime,chshop,chm);}
gotoxy(1,23);printf("|----------------------------------------------|");
gotoxy(1,24);printf("| |");
gotoxy(1,25);printf("|----------------------------------------------|");
gotoxy(10,24);printf("total is %8.1f$",sum);
fclose(fp);
gotoxy(49,24);printf("press any key to.....");getch();goto pp;
}
else
{
while(ch[0]!='\r')
{
if(j10)
{
strncat(chtime,ch,1);
j++;
}
if(ch[0]==8)
{
len=strlen(chtime)-1;
if(j15)
{
len=len+1;
j=11;
}
strcpy(ch1,"");
j=j-2;
strncat(ch1,chtime,len);
strcpy(chtime,"");
strncat(chtime,ch1,len-1);
gotoxy(13,7);printf(" ");
}
gotoxy(13,7);printf("%s",chtime);ch[0]=getch();
if(ch[0]==9)
goto mm;
if(ch[0]==27)
exit(1);
}
gotoxy(3,24);printf(" ");
gotoxy(13,10);
j=0;
ch[0]=getch();
while(ch[0]!='\r')
{
if (j14)
{
strncat(chshop,ch,1);
j++;
}
if(ch[0]==8)
{
len=strlen(chshop)-1;
strcpy(ch1,"");
j=j-2;
strncat(ch1,chshop,len);
strcpy(chshop,"");
strncat(chshop,ch1,len-1);
gotoxy(13,10);printf(" ");
}
gotoxy(13,10);printf("%s",chshop);ch[0]=getch();}
gotoxy(13,13);
j=0;
ch[0]=getch();
while(ch[0]!='\r')
{
if (j6)
{
strncat(chmoney,ch,1);
j++;
}
if(ch[0]==8)
{
len=strlen(chmoney)-1;
strcpy(ch1,"");
j=j-2;
strncat(ch1,chmoney,len);
strcpy(chmoney,"");
strncat(chmoney,ch1,len-1);
gotoxy(13,13);printf(" ");
}
gotoxy(13,13);printf("%s",chmoney);ch[0]=getch();
}
if((strlen(chshop)==0)||(strlen(chmoney)==0))
continue;
if((fp=fopen("home.dat","a+"))!=NULL);
fprintf(fp,"%10s%14s%6s",chtime,chshop,chmoney);
fputc('\n',fp);
fclose(fp);
i++;
gotoxy(41,5+i);
printf("%10s %-14s %-6s",chtime,chshop,chmoney);
}
}
}
【程序96】
题目：计算字符串中子串出现的次数
1.程序分析：
2.程序源代码：
#include "string.h"
#include "stdio.h"
main()
{
char str1[20],str2[20],*p1,*p2;
int sum=0;
printf("please input two strings\n");
scanf("%s%s",str1,str2);
p1=str1;p2=str2;
while(*p1!='\0')
{
if(*p1==*p2)
{
while(*p1==*p2*p2!='\0')
{
p1++;
p2++;
}
}
else
p1++ ;
if(*p2=='\0')
sum++;
p2=str2;
}
printf("%d",sum);
getch();
}
【程序97】
题目：从键盘输入一些字符，逐个把它们送到磁盘上去，直到输入一个#为止。
1.程序分析：
2.程序源代码：
#include "stdio.h"
main()
{
FILE *fp;
char ch,filename[10];
scanf("%s",filename);
if((fp=fopen(filename,"w"))==NULL)
{
printf("cannot open file\n");
exit(0);
}
ch=getchar();
ch=getchar();
while(ch!='#')
{
fputc(ch,fp);putchar(ch);
ch=getchar();
}
fclose(fp);
}
【程序98】
题目：从键盘输入一个字符串，将小写字母全部转换成大写字母，然后输出到一个磁盘文件"test"中保存。输入的字符串以！结束。
1.程序分析：
2.程序源代码：
#include "stdio.h"
main()
{
FILE *fp;
char str[100],filename[10];
int i=0;
if((fp=fopen("test","w"))==NULL)
{
printf("cannot open the file\n");
exit(0);
}
printf("please input a string:\n");
gets(str);
while(str[i]!='!')
{
if(str[i]='a'str[i]='z')
str[i]=str[i]-32;
fputc(str[i],fp);
i++;
}
fclose(fp);
fp=fopen("test","r");
fgets(str,strlen(str)+1,fp);
printf("%s\n",str);
fclose(fp);
}
【程序99】
题目：有两个磁盘文件A和B,各存放一行字母，要求把这两个文件中的信息合并（按字母顺序排列），
输出到一个新文件C中。
1.程序分析：
2.程序源代码：
#include "stdio.h"
main()
{
FILE *fp;
int i,j,n,ni;
char c[160],t,ch;
if((fp=fopen("A","r"))==NULL)
{
printf("file A cannot be opened\n");
exit(0);
}
printf("\n A contents are :\n");
for(i=0;(ch=fgetc(fp))!=EOF;i++)
{
c[i]=ch;
putchar(c[i]);
}
fclose(fp);
ni=i;
if((fp=fopen("B","r"))==NULL)
{
printf("file B cannot be opened\n");
exit(0);
}
printf("\n B contents are :\n");
for(i=0;(ch=fgetc(fp))!=EOF;i++)
{
c[i]=ch;
putchar(c[i]);
}
fclose(fp);
n=i;
for(i=0;in;i++)
for(j=i+1;jn;j++)
if(c[i]c[j])
{
t=c[i];c[i]=c[j];c[j]=t;
}
printf("\n C file is:\n");
fp=fopen("C","w");
for(i=0;in;i++)
{
putc(c[i],fp);
putchar(c[i]);
}
fclose(fp);
}
【程序100】
题目：有五个学生，每个学生有3门课的成绩，从键盘输入以上数据（包括学生号，姓名，三门课成绩），计算出平均成绩，况原有的数据和计算出的平均分数存放在磁盘文件"stud"中。
1.程序分析：
2.程序源代码：
#include "stdio.h"
struct student
{
char num[6];
char name[8];
int score[3];
float avr;
} stu[5];
main()
{
int i,j,sum;
FILE *fp;
/*input*/
for(i=0;i5;i++)
{
printf("\n please input No. %d score:\n",i);
printf("stuNo:");
scanf("%s",stu[i].num);
printf("name:");
scanf("%s",stu[i].name);
sum=0;
for(j=0;j3;j++)
{
printf("score %d.",j+1);
scanf("%d",stu[i].score[j]);
sum+=stu[i].score[j];
}
stu[i].avr=sum/3.0;
}
fp=fopen("stud","w");
for(i=0;i5;i++)
if(fwrite(stu[i],sizeof(struct student),1,fp)!=1)
printf("file write error\n");
fclose(fp);
}
c语言编程:找出200以内所有完数,并输出其因子 输入输出示例：1=2 6=1+2+3 28=1+2+4+7+14
C语言程序：
#include "stdio.h"
void main()
{
int n, sum;
int i;
for(n=1; n=200; n++)
{
sum = 0;
for(i=1; in; i++)
if(n % i == 0)
sum += i;
if(sum == n)
{
printf("%d=1", n);
for(i=2; in; i++)
if(n % i == 0)
printf("+%d", i);
printf("\n");
}
}
}
运行结果：
6=1+2+3
28=1+2+4+7+14
c语言编程代码
两种方法我写在一起，可以独立拆开。
#include stdio.h
void finda1(char a[3][10]);
void finda2(char a[3][10]);
void show(char (*p)[10]);
int main()
{
char a[3][10]={{"gehajl"},{"788a987a7"},{"ccabbbabbb"}};
printf("原数组内容:\n");
show(a);
printf("\n1、用数组指针的方法（函数finda1）：\n");
finda1(a);
printf("执行后:\n");
show(a);
printf("\n---------------------\n");
char b[3][10]={{"gehajl"},{"788a987a7"},{"ccabbbabbb"}};
printf("原数组内容:\n");
show(a);
printf("\n2、用指针数组的方法（函数finda2）：\n");
finda2(b);
printf("执行后:\n");
show(b);
return 0;
}
void finda1(char a[3][10])
{
int i,j;
char (*p)[10]=a;
for(i=0;i3;i++)
for(j=0;j10;j++)
if(p[i][j]=='a')
printf("发现：第%d行第%d个元素是‘a’，已替换\n",i+1,j+1),p[i][j]='1';
}
void finda2(char a[3][10])
{
int i,j;
char *p[3]={a[0][0],a[1][0],a[2][0]};
for(i=0;i3;i++)
for(j=0;j10;j++)
if(p[i][j]=='a')
printf("发现：第%d行第%d个元素是‘a’，已替换\n",i+1,j+1),p[i][j]='1';
}
void show(char (*p)[10])
{
int i,j;
for(i=0;i3;i++,printf("\n"))
for(j=0;j10;j++)
printf("%c ",p[i][j]);
}
