经典程序100例（91-100）

　　　　经典c程序100例==81--90【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); }}==============================================================【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);}==============================================================【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;elses*=8;sum+=s;}printf("\nsum=%ld",sum);}==============================================================【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;elsebreak;for(c=2;c<=sqrt(d);c++)if(d%c==0) break;if(c>sqrt(d))printf("%d=%d+%d\n",a,b,d);}}==============================================================【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);}==============================================================【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++;}elsec[k]=b[j++];k++;}c[k]='\0';if(a[i]=='\0')p=b+j;elsep=a+i;strcat(c,p);puts(c);}==============================================================【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';}==============================================================【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();}==============================================================【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]);}==============================================================【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 5main(){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));}

经典c程序100例==91--100【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);/*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21.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31.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 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");elseprintf("\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);}elseif((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++;}}elsep1++;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;ic[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