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From: =?gb2312?B?08kgV2luZG93cyBJbnRlcm5ldCBFeHBsb3JlciA3ILGjtOY=?=
Subject: 1001 -- Exponentiation
Date: Sat, 27 Dec 2008 04:36:09 +0800
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    <TH class=3Dh>Online Judge</TH>
    <TH class=3Dh>Problem Set</TH>
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onchange=3D"window.location.href=3D'problem?id=3D1001&amp;lang=3D'+this.v=
alue+'&amp;change=3Dtrue'"=20
      size=3D1><OPTION value=3Ddefault selected>Default</OPTION><OPTION=20
        =
value=3Dzh-CN>=E7=AE=80=E4=BD=93=E4=B8=AD=E6=96=87</OPTION></SELECT></DIV=
>
      <DIV class=3Dptt lang=3Den-US>Exponentiation</DIV>
      <DIV class=3Dplm>
      <TABLE align=3Dcenter>
        <TBODY>
        <TR>
          <TD><B>Time Limit:</B> 500MS</TD>
          <TD width=3D10></TD>
          <TD><B>Memory Limit:</B> 10000K</TD></TR>
        <TR>
          <TD><B>Total Submissions:</B> 43976</TD>
          <TD width=3D10></TD>
          <TD><B>Accepted:</B> 10083</TD></TR></TBODY></TABLE></DIV>
      <P class=3Dpst>Description</P>
      <DIV class=3Dptx lang=3Den-US>Problems involving the computation =
of exact=20
      values of very large magnitude and precision are common. For =
example, the=20
      computation of the national debt is a taxing experience for many =
computer=20
      systems. <BR><BR>This problem requires that you write a program to =
compute=20
      the exact value of R<SUP>n</SUP> where R is a real number ( 0.0 =
&lt; R=20
      &lt; 99.999 ) and n is an integer such that 0 &lt; n &lt;=3D 25. =
</DIV>
      <P class=3Dpst>Input</P>
      <DIV class=3Dptx lang=3Den-US>The input will consist of a set of =
pairs of=20
      values for R and n. The R value will occupy columns 1 through 6, =
and the n=20
      value will be in columns 8 and 9.</DIV>
      <P class=3Dpst>Output</P>
      <DIV class=3Dptx lang=3Den-US>The output will consist of one line =
for each=20
      line of input giving the exact value of R^n. Leading zeros should =
be=20
      suppressed in the output. Insignificant trailing zeros must not be =

      printed. Don't print the decimal point if the result is an =
integer.</DIV>
      <P class=3Dpst>Sample Input</P><PRE class=3Dsio>95.123 12
0.4321 20
5.1234 15
6.7592  9
98.999 10
1.0100 12
</PRE>
      <P class=3Dpst>Sample Output</P><PRE =
class=3Dsio>548815620517731830194541.899025343415715973535967221869852721=

.000000051485546410769561219945112767671548384817602007263512038354297630=
13462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201</PRE>
      <P class=3Dpst>Hint</P>
      <DIV class=3Dptx lang=3Den-US>If you don't know how to determine =
wheather=20
      encounted the end of input: <BR><I>s</I> is a string and <I>n</I> =
is an=20
      integer <BR><PRE><B>C++</B>
<BR>while(cin&gt;&gt;s&gt;&gt;n)
<BR>{
<BR>...
<BR>}
<BR><B>c</B>
<BR>while(scanf("%s%d",s,&amp;n)=3D=3D2) //to  see if the scanf read in =
as many items as you want
<BR>/*while(scanf(%s%d",s,&amp;n)!=3DEOF) //this also work    */
<BR>{
<BR>...
<BR>}</PRE></DIV>
      <P class=3Dpst>Source</P>
      <DIV class=3Dptx lang=3Den-US><A=20
      =
href=3D"http://acm.pku.edu.cn/JudgeOnline/searchproblem?field=3Dsource&am=
p;key=3DEast+Central+North+America+1988">East=20
      Central North America =
1988</A></DIV></TD></TR></TBODY></TABLE><FONT=20
color=3D#333399 size=3D3>
<P align=3Dcenter>[<A=20
href=3D"http://acm.pku.edu.cn/JudgeOnline/submit?problem_id=3D1001">Submi=
t</A>]&nbsp;&nbsp;=20
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