c语言之辗转相除法求最大公约数

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</pre><pre name="code" class="html"></pre>辗转相除法</h1><p><span style="font-size:14px">      <span style="color:rgb(51,51,51); font-family:arial,宋体,sans-serif; font-size:14px; line-height:24px; text-indent:28px">辗转相除法，又名</span><a target=_blank target="_blank" href="http://baike.baidu.com/view/1241014.htm" style="text-decoration:none; color:rgb(19,110,194); font-family:arial,宋体,sans-serif; font-size:14px; line-height:24px; text-indent:28px">欧几里德算法</a><span style="color:rgb(51,51,51); font-family:arial,宋体,sans-serif; font-size:14px; line-height:24px; text-indent:28px">（Euclidean algorithm）乃求两个</span><a target=_blank target="_blank" href="http://baike.baidu.com/view/464125.htm" style="text-decoration:none; color:rgb(19,110,194); font-family:arial,宋体,sans-serif; font-size:14px; line-height:24px; text-indent:28px">正整数</a><span style="color:rgb(51,51,51); font-family:arial,宋体,sans-serif; font-size:14px; line-height:24px; text-indent:28px">之</span><a target=_blank target="_blank" href="http://baike.baidu.com/view/1103370.htm" style="text-decoration:none; color:rgb(19,110,194); font-family:arial,宋体,sans-serif; font-size:14px; line-height:24px; text-indent:28px">最大公因子</a><span style="color:rgb(51,51,51); font-family:arial,宋体,sans-serif; font-size:14px; line-height:24px; text-indent:28px">的算法。它是已知最古老的算法， 其可追溯至3000年前。</span></span></p><p><span style="font-size:14px"><span style="color:rgb(51,51,51); font-family:arial,宋体,sans-serif; font-size:14px; line-height:24px; text-indent:28px">       其过程为：已知a，b两个数，其中a大于b。则a对b求于a%b，得到余数r。因为余数r小于b，则把b，r中较大的数b赋值给a，r赋值给b，再b对r求余数，以此往复，直到余数为零。</span></span></p><p style="text-indent:28px"><span style="font-family:arial,宋体,sans-serif; color:#333333"><span style="font-size:14px; line-height:24px">在下面辗转相除法函数创建为：void Euclid(int a, int b);</span></span></p><h1><span style="line-height:24px; color:rgb(51,51,51); font-family:arial,宋体,sans-serif; font-size:24px">原理</span></h1><p style="text-indent:28px"><span style="font-family:arial,宋体,sans-serif; color:#333333"><span style="font-size:14px; line-height:24px"></span></span></p><div class="para" style="color:rgb(51,51,51); margin:15px 0px 5px; text-indent:2em; line-height:24px; font-family:arial,宋体,sans-serif; font-size:14px">设两数为a、b(b<a），用gcd(a,b）表示a，b的最大公约数，r=a mod b 为a除以b以后的余数，k为a除以b的商，即a÷b=k.......r。辗转相除法即是要证明gcd(a,b)=gcd(b,r）。</div><div class="para" style="color:rgb(51,51,51); margin:15px 0px 5px; text-indent:2em; line-height:24px; font-family:arial,宋体,sans-serif; font-size:14px">第一步：令c=gcd(a,b），则设a=mc，b=nc</div><div class="para" style="color:rgb(51,51,51); margin:15px 0px 5px; text-indent:2em; line-height:24px; font-family:arial,宋体,sans-serif; font-size:14px">第二步：根据前提可知r =a-kb=mc-knc=(m-kn)c</div><div class="para" style="color:rgb(51,51,51); margin:15px 0px 5px; text-indent:2em; line-height:24px; font-family:arial,宋体,sans-serif; font-size:14px">第三步：根据第二步结果可知c也是r的因数</div><div class="para" style="color:rgb(51,51,51); margin:15px 0px 5px; text-indent:2em; line-height:24px; font-family:arial,宋体,sans-serif; font-size:14px">第四步：可以断定m-kn与n互质【否则，可设m-kn=xd，n=yd，（d>1），则m=kn+xd=kyd+xd=(ky+x)d，则a=mc=(ky+x)dc，b=nc=ycd，故a与b最大公约数成为cd，而非c，与前面结论矛盾】</div><div class="para" style="color:rgb(51,51,51); margin:15px 0px 5px; text-indent:2em; line-height:24px; font-family:arial,宋体,sans-serif; font-size:14px">（互质：公约数只有1的两个数，称为互质数）</div><div class="para" style="color:rgb(51,51,51); margin:15px 0px 5px; text-indent:2em; line-height:24px; font-family:arial,宋体,sans-serif; font-size:14px">从而可知gcd(b,r)=c，继而gcd(a,b)=gcd(b,r）。</div><div class="para" style="color:rgb(51,51,51); margin:15px 0px 5px; text-indent:2em; line-height:24px; font-family:arial,宋体,sans-serif; font-size:14px">证毕。</div><h1>算法代码如下：</h1><div></div><pre name="code" class="objc" style="font-size: 24px; font-weight: bold;">void euclid(int a,int b){    int temp;        if(a < b)    {        temp = a;        a = b;        b = temp;    }        while(temp=a%b)    {        a = b;        b = temp;    }        printf("a和b的最大公约数为：%d",b);    }