p=s;
gets(p);
n=0;
while(*(p)!='\0')
{n=n*8+*p-'0';
p++;}
printf("%d",n);
}
------------------------------------------------------------------------------
http://www.js0573.com/yx/jsfs/4509220.html
http://www.zznews.cn/hyzx/jkxx/5469347888.html
http://www.rznews.cn/yszx/stbk/5491478666.html
http://www.yltvb.com/xwbk/jkxx/5464296969.html
http://www.rznews.cn/yszx/stbk/5487556791.html
http://www.zznews.cn/hyzx/jkxx/5464185928.html
http://www.rznews.cn/yszx/stbk/5487032076.html
http://www.lfxww.com/jk/zxzx/5456187661.html
http://www.rznews.cn/yszx/stbk/5479584797.html
http://www.yltvb.com/xwbk/jkxx/5461338233.html
http://www.lfxww.com/jk/zxzx/5456190414.html
http://www.yltvb.com/xwbk/jkxx/5461338233.html
http://www.rznews.cn/yszx/stbk/5479584797.html
http://www.lfxww.com/jk/zxzx/5456188652.html
http://www.yltvb.com/xwbk/jkxx/5461343939.html
http://www.zznews.cn/hyzx/jkxx/5456688226.html
http://www.lfxww.com/jk/zxzx/5456187661.html
http://www.zznews.cn/hyzx/jkxx/5456694669.html
http://www.yltvb.com/xwbk/jkxx/5461376849.html
http://www.zznews.cn/hyzx/jkxx/5464170704.html
http://yiyuan.hangzhou.com.cn/hmfx/4485699.html
http://www.rznews.cn/yszx/stbk/5487032076.html
http://yiyuan.hangzhou.com.cn/hmfx/4485690.html
http://www.zznews.cn/hyzx/jkxx/5464185928.html
http://yiyuan.hangzhou.com.cn/hmfx/4485682.html
http://www.rznews.cn/yszx/stbk/5487556791.html
http://yiyuan.hangzhou.com.cn/hmfx/4485705.html
http://www.zznews.cn/hyzx/jkxx/5464191155.html
http://www.yltvb.com/xwbk/jkxx/5464296969.html
http://yiyuan.hangzhou.com.cn/hmfx/4485701.html
http://www.zznews.cn/hyzx/jkxx/5468442672.html
http://yiyuan.hangzhou.com.cn/hmfx/4486883.html
http://www.rznews.cn/yszx/stbk/5491478666.html
http://www.zznews.cn/hyzx/jkxx/5468519681.html
http://yiyuan.hangzhou.com.cn/hmfx/4486890.html
http://www.qdxw.com.cn/yx/dwwf/2525169.html
http://www.zznews.cn/hyzx/jkxx/5469347888.html
http://www.zznews.cn/hyzx/jkxx/5468554744.html
http://yiyuan.hangzhou.com.cn/hmfx/4486893.html
http://www.js0573.com/yx/jsfs/4509220.html
【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;
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()
{ 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;
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);
}
}
-----------------------------------------------------------------------------