?? 程序員考試模擬試卷——下午題[3] - 學(xué)賽網(wǎng)軟考學(xué)院.htm
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<P> 本程序?qū)㈦姶a的原文存放在字符數(shù)組 old 中，加密鑰匙存放在整數(shù) key 中。函數(shù) decode 用于將原文 old
加密并返回密文字符數(shù)組的首指針。其中函數(shù)采用一個(gè)雙向循環(huán)鏈表 CODE 來(lái)表示密文環(huán).函數(shù) strlen 用于計(jì)算一個(gè)字符串中的字符個(gè)數(shù)( 不包括字符串結(jié)尾符
'\0' )。為了簡(jiǎn)單起見(jiàn)，程序中假設(shè)內(nèi)存容量足以滿(mǎn)足動(dòng)態(tài)存貯單元分配的要求。</P>
<P> #include <STDIO.H></P>
<P> #include <STDLIB.H></P>
<P> typedef struct node</P>
<P> { char ch;</P>
<P> struct node *forward; /* Link to next node. */</P>
<P> struct node *backward;/* Link to previous node.*/</P>
<P> } CODE;</P>
<P> int strlen(char *s)</P>
<P> { int len = 0;</P>
<P> while (*s++ != '\0' )</P>
<P> len++;</P>
<P> return( len );</P>
<P> }</P>
<P> char *decode(char *old,int key)</P>
<P> { char *New; int length,count,i;</P>
<P> CODE *loop,*p;</P>
<P> length=strlen(old);</P>
<P> loop=(CODE *) malloc( length*sizeof(CODE) );</P>
<P> for ( i = 1;i<LENGTH-1;I++ p )<>
<P> { loop[i].forward = &loop[i+1];</P>
<P> ___(1)___</P>
<P> }</P>
<P> loop[0].backward = &loop[length-1];</P>
<P> loop[0].forward = &loop[1];</P>
<P> loop[length-1].forward = loop;</P>
<P> ___(2)___</P>
<P> for ( p = loop,i = 0;i<LENGTH;I++ p )<>
<P> { for ( count = 1;count<KEY;COUNT++ p )<>
<P> p= p->forward ;</P>
<P> ___(3)___</P>
<P> p->backward->forward = p->forward ;</P>
<P> p->forward->backward = p->backward ;</P>
<P> ___(4)___</P>
<P> }</P>
<P> New = ( char *)malloc( ( length+1 ) *sizeof(char) );</P>
<P> for ( i=0;i<LENGTH;I++ p )<>
<P> ___(5)___</P>
<P> New[length]='\0';</P>
<P> return (New);</P>
<P> }</P>
<P> void main()</P>
<P> { char old[256];</P>
<P></P>
<P></P>
<P></P>
<P align=right>[<A
href="http://www.educity.cn/rk/prog/200804111009281842.htm">1</A>] [<A
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[<A href="http://www.educity.cn/rk/prog/200804111009281140.htm">4</A>]
[<A href="http://www.educity.cn/rk/prog/200804111009291173.htm">5</A>]
[<A href="http://www.educity.cn/rk/prog/200804111009291137.htm">7</A>]
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