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2023年12月18日发(作者:数据库试题)
经典C源代码30例
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【程序80】
题目:海滩上有一堆桃子,五只猴子来分。第一只猴子把这堆桃子凭据分为五份,多了一个,这只
猴子把多的一个扔入海中,拿走了一份。第二只猴子把剩下的桃子又平均分成五份,又多了
一个,它同样把多的一个扔入海中,拿走了一份,第三、第四、第五只猴子都是这样做的,
问海滩上原来最少有多少个桃子?
1.程序分析:
2.程序源代码:
main()
{int i,m,j,k,count;
for(i=4;i<10000;i+=4)
{ count=0;
m=i;
for(k=0;k<5;k++)
{
j=i/4*5+1;
i=j;
if(j%4==0)
count++;
else
break;
}
i=m;
if(count==4)
{printf("%dn",count);
break;}
}
}
作者: zhlei81 2005-1-22 11:32 回复此发言
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16 回复:经典C源程序100例
【程序81】
题目:809*??=800*??+9*??+1 其中??代表的两位数,8*??的结果为两位数,9*??的结果为3位数。求??代表的两位数,及809*??后的结果。
1.程序分析:
2.程序源代码:
output(long b,long i)
{ printf("n%ld/%ld=809*%ld+%ld",b,i,i,b%i);
}
main()
{long int a,b,i;
a=809;
for(i=10;i<100;i++)
{b=i*a+1;
if(b>=1000&&b<=10000&&8*i<100&&9*i>=100)
output(b,i); }
}
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【程序82】
题目:八进制转换为十进制
1.程序分析:
2.程序源代码:
main()
{ char *p,s[6];int n;
p=s;
gets(p);
n=0;
while(*(p)!='0')
{n=n*8+*p-'0';
p++;}
printf("%d",n);
}
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【程序83】
题目:求0—7所能组成的奇数个数。
1.程序分析:
2.程序源代码:
main()
{
long sum=4,s=4;
int j;
for(j=2;j<=8;j++)/*j is place of number*/
{ printf("n%ld",sum);
if(j<=2)
s*=7;
else
s*=8;
sum+=s;}
printf("nsum=%ld",sum);
}
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【程序84】
题目:一个偶数总能表示为两个素数之和。
1.程序分析:
2.程序源代码:
#include "stdio.h"
#include "math.h"
main()
{ int a,b,c,d;
scanf("%d",&a);
for(b=3;b<=a/2;b+=2)
{ for(c=2;c<=sqrt(b);c++)
if(b%c==0) break;
if(c>sqrt(b))
d=a-b;
else
break;
for(c=2;c<=sqrt(d);c++)
if(d%c==0) break;
if(c>sqrt(d))
printf("%d=%d+%dn",a,b,d);
}
}
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【程序85】
题目:判断一个素数能被几个9整除
1.程序分析:
2.程序源代码:
main()
{ long int m9=9,sum=9;
int zi,n1=1,c9=1;
scanf("%d",&zi);
while(n1!=0)
{ if(!(sum%zi))
n1=0;
else
{m9=m9*10;
sum=sum+m9;
c9++;
}
}
printf("%ld,can be divided by %d "9"",sum,c9);
}
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【程序86】
题目:两个字符串连接程序
1.程序分析:
2.程序源代码:
#include "stdio.h"
main()
{char a[]="acegikm";
char b[]="bdfhjlnpq";
char c[80],*p;
int i=0,j=0,k=0;
while(a[i]!='0'&&b[j]!='0')
{if (a[i] { c[k]=a[i];i++;}
else
c[k]=b[j++];
k++;
}
c[k]='0';
if(a[i]=='0')
p=b+j;
else
p=a+i;
strcat(c,p);
puts©;
}
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【程序87】
题目:回答结果(结构体变量传递)
1.程序分析:
2.程序源代码:
#include "stdio.h"
struct student
{ int x;
char c;
} a;
main()
{a.x=3;
a.c='a';
f(a);
printf("%d,%c",a.x,a.c);
}
f(struct student b)
{
b.x=20;
b.c='y';
}
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【程序88】
题目:读取7个数(1—50)的整数值,每读取一个值,程序打印出该值个数的*。
1.程序分析:
2.程序源代码:
main()
{int i,a,n=1;
while(n<=7)
{ do {
scanf("%d",&a);
}while(a<1¦¦a>50);
for(i=1;i<=a;i++)
printf("*");
printf("n");
n++;}
getch();
}
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【程序89】
题目:某个公司采用公用电话传递数据,数据是四位的整数,在传递过程中是加密的,加密规则如下:
每位数字都加上5,然后用和除以10的余数代替该数字,再将第一位和第四位交换,第二位和第三位交换。
1.程序分析:
2.程序源代码:
main()
{int a,i,aa[4],t;
scanf("%d",&a);
aa[0]=a%10;
aa[1]=a%100/10;
aa[2]=a%1000/100;
aa[3]=a/1000;
for(i=0;i<=3;i++)
{aa[i]+=5;
aa[i]%=10;
}
for(i=0;i<=3/2;i++)
{t=aa[i];
aa[i]=aa[3-i];
aa[3-i]=t;
}
for(i=3;i>=0;i--)
printf("%d",aa[i]);
}
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【程序90】
题目:专升本一题,读结果。
1.程序分析:
2.程序源代码:
#include "stdio.h"
#define M 5
main()
{int a[M]={1,2,3,4,5};
int i,j,t;
i=0;j=M-1;
while(i {t=*(a+i);
*(a+i)=*(a+j);
*(a+j)=t;
i++;j--;
}
for(i=0;i printf("%d",*(a+i));
}
作者: zhlei81 2005-1-22 11:33 回复此发言
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17 回复:经典C源程序100例
【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*/
}
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【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("1111111111n");}
end=time(NULL);
printf("1: The different is %6.3fn",difftime(end,start));
}
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【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("1111111111n");}
end=clock();
printf("1: The different is %6.3fn",(double)(end-start));
}
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【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(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 secondsn",var=(double)(end-start)/18.2);
printf("1: it took you %6.3f secondsnn",difftime(b,a));
if(var<15)
printf("11 You are very clever! 11nn");
else if(var<25)
printf("11 you are normal! 11nn");
else
printf("11 you are stupid! 11nn");
printf("11 Congradulations 11nn");
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;
}
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【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 --
¦");
作者: zhlei81 2005-1-22 11:33 回复此发言
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18 回复:经典C源程序100例
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",_year,_mon,_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("","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%fn",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 %6.1fn",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 ");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("","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);
}}}
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【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 stringsn");
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();}
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【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 filen");
exit(0);}
ch=getchar();
ch=getchar();
while(ch!='#')
{fputc(ch,fp);putchar(ch);
ch=getchar();
}
fclose(fp);
}
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【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 filen");
exit(0);}
printf("please input a string:n");
gets(str);
while(str!='!')
{ if(str>='a'&&str<='z')
str=str-32;
fputc(str,fp);
i++;}
fclose(fp);
fp=fopen("test","r");
fgets(str,strlen(str)+1,fp);
printf("%sn",str);
fclose(fp);
}
作者: zhlei81 2005-1-22 11:33 回复此发言
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19 回复:经典C源程序100例
【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 openedn");
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 openedn");
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 for(j=i+1;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 { 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 errorn"); fclose(fp);
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