Google面试的一道题目及解答

qc1125

<span id="post1" style="FONT-SIZE: 12px; COLOR: #000000;">题目如下：<br/><br/>题目：解答下面的隐藏等式，其中的M和E的值可以互换，但不允许第一位是0。<br/>WWWDOT - GOOGLE = DOTCOM <br/><br/>解答代码：<br/>/*******************************/ <br/>/* &nbsp; Q:WWWDOT-GOOGLE=DOTCOM &nbsp;*/ <br/>/* &nbsp; A:777589-188103=589486 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;*/ <br/>/* &nbsp; &nbsp; 777589-188106=589483 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; */ <br/>/*******************************/ <br/><br/><br/>#include &lt;stdio.h&gt; <br/><br/>void getans(int num[], int n_tmp) &nbsp; &nbsp;//通过递归调用对num_0_9数组进行组合排序 <br/>{ <br/>&nbsp;int tmp; <br/>&nbsp;int w,d,o,t,g,l,e,c,m,wwwdot,google,dotcom; <br/>&nbsp;if(n_tmp == 1){ <br/>&nbsp; &nbsp;w = num[0]; <br/>&nbsp; &nbsp;d = num[1]; <br/>&nbsp; &nbsp;o = num[2]; <br/>&nbsp; &nbsp;t = num[3]; <br/>&nbsp; &nbsp;g = num[4]; <br/>&nbsp; &nbsp;l = num[5]; <br/>&nbsp; &nbsp;e = num[6]; <br/>&nbsp; &nbsp;c = num[7]; <br/>&nbsp; &nbsp;m = num[8]; <br/>&nbsp; &nbsp;wwwdot = w*111000 + d*100 + o*10 + t; <br/>&nbsp; &nbsp;google = g*100100 + o*11000 + l*10 + e; <br/>&nbsp; &nbsp;dotcom = d*100000 + o*10010 + t*1000 + c*100 + m; <br/><br/>&nbsp; &nbsp;//其中的M和E的值可以互换，但不允许第一位是0。 <br/>&nbsp; &nbsp;if(w!=0 &amp;&amp; g!=0 &amp;&amp; d!=0) &nbsp; &nbsp; <br/>&nbsp; &nbsp; &nbsp; &nbsp;if((wwwdot - google == dotcom) || (wwwdot - (google - e + m) == (dotcom - m + e))) <br/>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;printf("%d - %d = %d\n",wwwdot,google,dotcom); <br/>&nbsp;} <br/>&nbsp;else{ <br/>&nbsp; &nbsp;for(int i = 0; i &lt; n_tmp; i++){ <br/>&nbsp; &nbsp; &nbsp;tmp=num; num=num[n_tmp-1]; num[n_tmp-1]=tmp; <br/>&nbsp; &nbsp; &nbsp;getans(num, n_tmp-1); <br/>&nbsp; &nbsp; &nbsp;tmp=num; num=num[n_tmp-1]; num[n_tmp-1]=tmp; <br/><br/>&nbsp; &nbsp;} <br/>&nbsp;} <br/>} <br/><br/>void main() <br/>{ <br/>&nbsp;int n = 10; <br/>&nbsp;int num_0_9[]={0,1,2,3,4,5,6,7,8,9}; <br/><br/>&nbsp;getans(num_0_9, n); <br/><br/>} <br/></span>