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📁 解小数问题(c实现)
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          <TD align=left width="100%" height=21>1.小数问题
            <P><FONT 
color=#0000ff>问题</FONT>:正整数相除用小数表示,有限小数直接写出结果,循环小数写出循环节。</P>
            <P><FONT 
            color=#0000ff>算法</FONT>:本算法是根据计算正整数相除的运算规则而做的,需要判断的输出有:1.整数;2.有限小</P>
            <P>数;3.纯循环小数;4.非纯循环小数。注意:不存在无限不循环小数。</P>
            <P>如果计算出来的余数和商在以前出现过,就说明找到了循环节,因此以两个数组来存放每次运算</P>
            <P>后的余数和商。</P>
            <P>/*<BR>=======================================================================<BR>If 
            you input m and n, this program can tell you what m/n is.<BR>For 
            example :&nbsp;<BR>#1. 1/4=0.25<BR>#2. 7/6=1.1(6)<BR>#3. 
            3/7=0.(428571)<BR>=======================================================================<BR>*/<BR>#include 
            &lt;stdio.h&gt;<BR>#include &lt;conio.h&gt;<BR><BR>#define TRUE 
            1<BR>#define FALSE 0<BR><BR>void main()<BR>{<BR>&nbsp;unsigned int 
            i=0,j,t,s,m,n,k;<BR>&nbsp;unsigned int q[1000],r[1000];<BR>&nbsp;int 
            big,small=TRUE,flag=TRUE,limit=TRUE;<BR>&nbsp;scanf("%d/%d",&amp;m,&amp;n);<BR>&nbsp;t=m%n;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
            /* m div n 的余数 
            */<BR>&nbsp;big=(m-t)/n;<BR>&nbsp;if(t==0)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
            /* m可以被n所整除,只输出整数部分 */<BR>&nbsp;{<BR>&nbsp; 
            printf("%d\n",big);<BR>&nbsp; 
            small=FALSE;<BR>&nbsp;}<BR><BR>&nbsp;if(t&gt;0) 
            printf("%d.",big);&nbsp; /* m不能被n所整除,有小数部分,所以要打出小数点 
            */<BR>&nbsp;q[0]=big;<BR>&nbsp;r[0]=t;<BR>&nbsp;do<BR>&nbsp;{<BR>&nbsp; 
            t=(r[i]*10)%n;<BR>&nbsp; i++;<BR>&nbsp; 
            q[i]=(r[i-1]*10-t)/n;<BR>&nbsp; r[i]=t;<BR>&nbsp; 
            for(j=1;j&lt;i;j++) /* 寻找是否出现了相同的余数和商 */<BR>&nbsp; {<BR>&nbsp;&nbsp; 
            if(q[j]==q[i]&amp;&amp;r[j]==r[i])<BR>&nbsp;&nbsp; 
            {<BR>&nbsp;&nbsp;&nbsp;&nbsp; 
            flag=FALSE;<BR>&nbsp;&nbsp;&nbsp;&nbsp; 
            for(k=j;k&lt;i;k++)<BR>&nbsp;&nbsp;&nbsp;&nbsp; 
            if(q[k]!=0)<BR>&nbsp;&nbsp;&nbsp; 
            {<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; limit=FALSE; /* 非有限 
            */<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
            break;<BR>&nbsp;&nbsp;&nbsp;&nbsp; }<BR>&nbsp;&nbsp;&nbsp;&nbsp; 
            if(limit) /* 有限小数 */<BR>&nbsp;&nbsp;&nbsp; 
            {<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
            for(k=1;k&lt;i-1;k++)<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
            printf("%d",q[k]);<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
            printf("\n");<BR>&nbsp;&nbsp;&nbsp;&nbsp; }<BR>&nbsp;&nbsp;&nbsp; 
            if(!limit)&nbsp;<BR>&nbsp;&nbsp; 
            {&nbsp;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; for(k=1;k&lt;j;k++) /* 
            不是循环节的部分 */<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
            printf("%d",q[k]);&nbsp;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
            printf("(");&nbsp; /* 循环从此开始 */<BR>&nbsp;&nbsp;&nbsp;&nbsp; 
            for(k=j;k&lt;i;k++)<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
            printf("%d",q[k]); /* q[k]是循环节&nbsp; k=j……i-1 
            */<BR>&nbsp;&nbsp;&nbsp;&nbsp; printf(")\n"); /* 循环从此结束 
            */<BR>&nbsp;&nbsp;&nbsp; }<BR>&nbsp;&nbsp; }<BR>&nbsp; 
            }<BR>&nbsp;}<BR>&nbsp;while(flag&amp;&amp;small) 
            ;<BR>&nbsp;getch();<BR>}</P>
            <P><BR>程序:<A 
            href="http://vip.6to23.com/dcyu/TurboC/senior/mdivn.c"><FONT 
            color=#0000ff>mdivn.c</FONT></A><FONT 
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            &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
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