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📁 经典一百例(c语言版)
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  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>
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                    <li><font color=#0772B1>上篇文章</font>:<a href="list.asp?id=210">经典c程序100例==71--80</a> 
                      
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            1.<a href="list.asp?id=212" title="经典c程序100例==91--100" target=_top> 
            经典c程序100例==91--100
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            经典c程序100例==1--10
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            <a href=list.asp?id=212>经典c程序100例==91--100</a><br>
            
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            <a href=list.asp?id=210>经典c程序100例==71--80</a><br>
            
            <a href=list.asp?id=209>经典c程序100例==61--70</a><br>
            
            <a href=list.asp?id=208>经典c程序100例==51--60</a><br>
            
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    <td width="87%" align=center>&nbsp;&nbsp;&nbsp;&nbsp;辽宁省铁岭师专c语言研究中心 <u>杨志锋</u> 数学系 <u>杜博</u> &nbsp;&nbsp;&nbsp;&nbsp;</td>
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