程序81100.docx
- 文档编号:27003407
- 上传时间:2023-06-25
- 格式:DOCX
- 页数:17
- 大小:18.82KB
程序81100.docx
《程序81100.docx》由会员分享,可在线阅读,更多相关《程序81100.docx(17页珍藏版)》请在冰豆网上搜索。
程序81100
【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(longb,longi)
{printf("\n%ld/%ld=809*%ld+%ld",b,i,i,b%i);
}
main()
{longinta,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];intn;
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()
{
longsum=4,s=4;
intj;
for(j=2;j<=8;j++)/*jisplaceofnumber*/
{printf("\n%ld",sum);
if(j<=2)
s*=7;
else
s*=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()
{inta,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+%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()
{longintm9=9,sum=9;
intzi,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,canbedividedby%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()
{chara[]="acegikm";
charb[]="bdfhjlnpq";
charc[80],*p;
inti=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(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"
structstudent
{intx;
charc;
}a;
main()
{a.x=3;
a.c='a';
f(a);
printf("%d,%c",a.x,a.c);
}
f(structstudentb)
{
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()
{inti,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()
{inta,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"
#defineM5
main()
{inta[M]={1,2,3,4,5};
inti,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)); } 【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" voidmain() {time_tlt;/*definealonginttimevarible*/ lt=time(NULL);/*systemtimeanddate*/ printf(ctime(<));/*englishformatoutput*/ printf(asctime(localtime(<)));/*tranfertotm*/ printf(asctime(gmtime(<)));/*tranfertoGreenwichtime*/ } ----------------------------------------------------------------------------- 【程序92】 题目: 时间函数举例2 1.程序分析: 2.程序源代码: /*calculatetime*/ #include"time.h" #include"stdio.h" main() {time_tstart,end; inti; 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: Thedifferentis%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: /*calculatetime*/ #include"time.h" #include"stdio.h" main() {clock_tstart,end; inti; doublevar; start=clock(); for(i=0;i<10000;i++) {printf("\1\1\1\1\1\1\1\1\1\1\n");} end=clock(); printf("\1: Thedifferentis%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() {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\1Youareveryclever! \1\1\n\n"); elseif(var<25) printf("\1\1youarenormal! \1\1\n\n"); else printf("\1\1youarestupid! \1\1\n\n"); printf("\1\1Congradulations\1\1\n\n"); printf("Thenumberyouguessis%d",i); } printf("\ndoyouwanttotryitagain? (\"yy\".or.\"n\")\n"); if((c=getch())=='y') gotoloop; } ----------------------------------------------------------------------------- 【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: /*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); }}} ----------------------------------------------------------------------------- 【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() {charstr1[20],str2[20],*p1,*p2; intsum=0; printf("please
- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- 程序 81100