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<td height="40" align="center" bgcolor="eeeeee"><font style="font-size:16px"><b>经典c程序100例==81--90</b></font></td>
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<td align="right"><font color="#999999">来源：<font color="#CC0000">郴州人才网</font>
时间：2004年10月27日8:5</font></td>
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<td class=c><font id="zoom" style="font-size:14px"><P><FONT color=#990000>【程序81】</FONT><BR>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。<BR>1.程序分析：<BR>2.程序源代码：<BR>output(long b,long i)<BR>{ printf("\n%ld/%ld=809*%ld+%ld",b,i,i,b%i);<BR>}<BR>main()<BR>{long int a,b,i;<BR>a=809;<BR>for(i=10;i<100;i++)<BR>{b=i*a+1;<BR>if(b>=1000&&b<=10000&&8*i<100&&9*i>=100)<BR>output(b,i); }<BR>}<BR>==============================================================<BR><FONT color=#990000>【程序82】</FONT><BR>题目：八进制转换为十进制<BR>1.程序分析： <BR>2.程序源代码：<BR>main()<BR>{ char *p,s[6];int n;<BR>p=s;<BR>gets(p);<BR>n=0;<BR>while(*(p)!='\0')<BR>{n=n*8+*p-'0';<BR>p++;}<BR>printf("%d",n);<BR>}<BR>==============================================================<BR><FONT color=#990000>【程序83】</FONT><BR>题目：求0—7所能组成的奇数个数。<BR>1.程序分析：<BR>2.程序源代码：<BR>main()<BR>{<BR>long sum=4,s=4;<BR>int j;<BR>for(j=2;j<=8;j++)/*j is place of number*/<BR>{ printf("\n%ld",sum);<BR>if(j<=2)<BR>s*=7;<BR>else<BR>s*=8;<BR>sum+=s;}<BR>printf("\nsum=%ld",sum);<BR>}<BR>==============================================================<BR><FONT color=#990000>【程序84】</FONT><BR>题目：一个偶数总能表示为两个素数之和。<BR>1.程序分析：<BR>2.程序源代码：<BR>#include "stdio.h"<BR>#include "math.h"<BR>main()<BR>{ int a,b,c,d;<BR>scanf("%d",&a);<BR>for(b=3;b<=a/2;b+=2)<BR>{ for(c=2;c<=sqrt(b);c++)<BR>if(b%c==0) break;<BR>if(c>sqrt(b))<BR>d=a-b;<BR>else<BR>break;<BR>for(c=2;c<=sqrt(d);c++)<BR>if(d%c==0) break;<BR>if(c>sqrt(d))<BR>printf("%d=%d+%d\n",a,b,d);<BR>}<BR>}<BR>==============================================================<BR><FONT color=#990000>【程序85】</FONT><BR>题目：判断一个素数能被几个9整除<BR>1.程序分析：<BR>2.程序源代码：<BR>main()<BR>{ long int m9=9,sum=9;<BR>int zi,n1=1,c9=1;<BR>scanf("%d",&zi);<BR>while(n1!=0)<BR>{ if(!(sum%zi))<BR>n1=0;<BR>else<BR>{m9=m9*10;<BR>sum=sum+m9;<BR>c9++;<BR>}<BR>}<BR>printf("%ld,can be divided by %d \"9\"",sum,c9);<BR>}<BR>==============================================================<BR><FONT color=#990000>【程序86】</FONT><BR>题目：两个字符串连接程序<BR>1.程序分析：<BR>2.程序源代码：<BR>#include "stdio.h"<BR>main()<BR>{char a[]="acegikm";<BR>char b[]="bdfhjlnpq";<BR>char c[80],*p;<BR>int i=0,j=0,k=0;<BR>while(a[i]!='\0'&&b[j]!='\0')<BR>{if (a[i]<B[J])<BR> { c[k]=a[i];i++;}<BR>else<BR>c[k]=b[j++];<BR>k++;<BR>}<BR>c[k]='\0';<BR>if(a[i]=='\0')<BR>p=b+j;<BR>else<BR>p=a+i;<BR>strcat(c,p);<BR>puts(c);<BR>}<BR>==============================================================<BR><FONT color=#990000>【程序87】</FONT><BR>题目：回答结果（结构体变量传递）<BR>1.程序分析： <BR>2.程序源代码：<BR>#include "stdio.h"<BR>struct student<BR>{ int x;<BR>char c;<BR>} a;<BR>main()<BR>{a.x=3;<BR>a.c='a';<BR>f(a);<BR>printf("%d,%c",a.x,a.c);<BR>}<BR>f(struct student b)<BR>{<BR>b.x=20;<BR>b.c='y';<BR>}<BR>==============================================================<BR><FONT color=#990000>【程序88】</FONT><BR>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＊。<BR>1.程序分析：<BR>2.程序源代码：<BR>main()<BR>{int i,a,n=1;<BR>while(n<=7)<BR>{ do {<BR> scanf("%d",&a);<BR> }while(a<1||a>50);<BR>for(i=1;i<=a;i++)<BR> printf("*");<BR>printf("\n");<BR>n++;}<BR>getch();<BR>}<BR>==============================================================<BR><FONT color=#990000>【程序89】</FONT><BR>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：<BR> 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。<BR>1.程序分析：<BR>2.程序源代码：<BR>main()<BR>{int a,i,aa[4],t;<BR>scanf("%d",&a);<BR>aa[0]=a%10;<BR>aa[1]=a%100/10;<BR>aa[2]=a%1000/100;<BR>aa[3]=a/1000;<BR>for(i=0;i<=3;i++)<BR> {aa[i]+=5;<BR> aa[i]%=10;<BR> }<BR>for(i=0;i<=3/2;i++)<BR> {t=aa[i];<BR> aa[i]=aa[3-i];<BR> aa[3-i]=t;<BR> }<BR>for(i=3;i>=0;i--)<BR>printf("%d",aa[i]);<BR>}<BR>==============================================================<BR><FONT color=#990000>【程序90】</FONT><BR>题目：专升本一题，读结果。<BR>1.程序分析：<BR>2.程序源代码：<BR>#include "stdio.h"<BR>#define M 5<BR>main()<BR>{int a[M]={1,2,3,4,5};<BR>int i,j,t;<BR>i=0;j=M-1;<BR>while(i<J)<BR> {t=*(a+i);<BR>*(a+i)=*(a+j);<BR>*(a+j)=t;<BR>i++;j--;<BR>}<BR>for(i=0;i<M;I++)<BR> printf("%d",*(a+i));<BR>}</P><BR> </font><br>
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