C语言.docx
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- 上传时间:2023-07-22
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C语言.docx
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C语言
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"
structstudent
{charnum[6];
charname[8];
intscore[3];
floatavr;
}stu[5];
main()
{inti,j,sum;
FILE*fp;
/*input*/
for(i=0;i<5;i++)
{printf("\npleaseinputNo.%dscore:
\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(structstudent),1,fp)!
=1)
printf("filewriteerror\n");
fclose(fp);
}
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;
inti,j,n,ni;
charc[160],t,ch;
if((fp=fopen("A","r"))==NULL)
{printf("fileAcannotbeopened\n");
exit(0);}
printf("\nAcontentsare:
\n");
for(i=0;(ch=fgetc(fp))!
=EOF;i++)
{c[i]=ch;
putchar(c[i]);
}
fclose(fp);考试大-全国最大教育类网站(www.Examda。
com)
ni=i;
if((fp=fopen("B","r"))==NULL)
{printf("fileBcannotbeopened\n");
exit(0);}
printf("\nBcontentsare:
\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("\nCfileis: \n"); fp=fopen("C","w"); for(i=0;i {putc(c[i],fp); putchar(c[i]); } fclose(fp); } 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;采集者退散 charstr[100],filename[10]; inti=0; if((fp=fopen("test","w"))==NULL) {printf("cannotopenthefile\n"); exit(0);} printf("pleaseinputastring: \n"); gets(str);考试大-全国最大教育类网站(www.Examda。 com) 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); } 题目: 家庭财务管理小程序 1.程序分析: 2.程序源代码: /*moneymanagementsystem*/ #include"stdio.h" #include"dos.h" main() { FILE*fp; structdated; floatsum,chm=0.0; intlen,i,j=0; intc; charch[4]="",ch1[16]="",chtime[12]="",chshop[16],chmoney[8]; pp: clrscr(); sum=0.0; gotoxy(1,1);printf("|---------------------------------------------------------------------------|"); gotoxy(1,2);printf("|moneymanagementsystem(C1.0)2000.03|"); gotoxy(1,3);printf("|---------------------------------------------------------------------------|"); gotoxy(1,4);printf("|--moneyrecords--|--todaycostlist--|"); 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__browsecostlistEsc__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("listrecords"); 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;} 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() {charstr1[20],str2[20],*p1,*p2; intsum=0; printf("pleaseinputtwostrings\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();} 题目: 家庭财务管理小程序 1.程序分析: 2.程序源代码: /*moneymanagementsystem*/ #include"stdio.h" #include"dos.h" main() { FILE*fp; structdated; floatsum,chm=0.0; intlen,i,j=0; intc; charch[4]="",ch1[16]="",chtime[12]="",chshop[16],chmoney[8]; pp: clrscr(); sum=0.0; gotoxy(1,1);printf("|---------------------------------------------------------------------------|"); gotoxy(1,2);printf("|moneymanagementsystem(C1.0)2000.03|"); gotoxy(1,3);printf("|---------------------------------------------------------------------------|"); gotoxy(1,4);printf("|--moneyrecords--|--todaycostlist--|"); 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__browsecostlistEsc__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("listrecords"); 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("totalis%8.1f$",sum); fclose(fp); gotoxy(49,24);printf("pressanykeyto.....");getch();gotopp; } 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) gotomm; 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); }}} 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() {charc; clock_tstart,end; time_ta,b; doublevar; inti,guess; srand(time(NULL)); printf("doyouwanttoplayit.('y'or'n')\n"); loop: while((c=getchar())=='y') { i=rand()%100; printf("\npleaseinputnumberyouguess: \n"); start=clock(); a=time(NULL); scanf("%d",&guess); while(guess! =i) {if(guess>i) {printf("pleaseinputalittlesmaller.\n"); scanf("%d",&guess);} else {printf("pleaseinputalittlebigger.\n"); scanf("%d",&guess);} } end=clock(); b=time(NULL); printf("\1: Ittookyou%6.3fseconds\n",var=(double)(end-start)/18.2); printf("\1: ittookyou%6.3fseconds\n\n",difftime(b,a)); if(var<15) printf("\1\1Youar
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