C语言程序设计经典实例之十
2008-03-08 22:03:12 来源:WEB开发网核心提示:【程序91】题目:时间函数举例11.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);/*sy
【程序91】
题目:时间函数举例1
1.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;i<3000;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;i<10000;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,判定一个人反应快慢。(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(guess>i)
{
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(var<15)
printf("\1\1 You are very clever! \1\1\n\n");
else if(var<25)
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;
} 更多内容请看C/C++进阶技术文档专题,或
【程序95】
题目:家庭财务治理小程序
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(j<10)
{
strncat(chtime,ch,1);
j++;
}
if(ch[0]==8)
{
len=strlen(chtime)-1;
if(j>15)
{
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 (j<14)
{
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 (j<6)
{
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);
} 更多内容请看C/C++进阶技术文档专题,或
【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;i<n;i++)
for(j=i+1;j<n;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;i<n;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;i<5;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;j<3;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;i<5;i++)
if(fwrite(&stu[i],sizeof(struct student),1,fp)!=1)
printf("file write error\n");
fclose(fp);
} 更多内容请看C/C++进阶技术文档专题,或
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