acm程序例题2道，还有一道思考的题目（第四题）

 1.Description   
  A   big   integer   with   most   of   its   digits   being   zeros   is   called   a   sparse   big   integer.   Given   a   sparse   big   integer   M   and   an   integer   N,   you   are   to   calculate   M   mod   N.     
    
  Input   
  The   first   line   contains   an   integer   representing   the   number   of   test   cases.   Each   test   case   consists   of   two   lines.   In   the   first   line   of   a   test   case,   an   integer   K   will   be   given   first;   then   K   pairs   of   integers   follow,   each   indicates   a   non-zero   digit   with   two   integers   D   and   P,   which   means   the   P-th   digit   of   M   from   right   to   left   is   D.   In   the   second   line,   the   integer   N   will   be   given.   
    
  1   <=   K   <=   10;   1   <=   D   <=   9;   1   <=   P   <=   1000000000;   1   <=   N   <=   10000.     
    
  Output   
  For   each   test   case,   print   the   result   of   M   mod   N   in   a   single   line.     
    
  Sample   Input     
    
  1   
  2   1   6   1   8   
  2046   
    
    
  Sample   Output     
    
  944   
    
  

//cpp文件   
  #include   "stdafx.h"   
  #include   "SparseBigInteger.h"   
  #include   <algorithm>   
  #include   <iostream>   
  #include   <map>   

    
  using   namespace   std;   
  static   int   loop_count   =   0;   
  long   lpow(long   v,   long   fa,   long   n)   
  {   
  map<long,   long>   series;   
  long   r   =   1;   
  long   se   =   0;   
  for   (long   f   =   0;   f   <   fa;   f++)   
  {   
  ++loop_count;   
  series[r]   =   se++;   
  r   *=   v;   
  r   %=   n;   
  if   (series.find(r)   !=   series.end())   
  {   
  r   *=   lpow(v,   (fa   -   series[r])   %   (se   -   series[r]),   n);   
  break;   
  }   
  }   
  return   r   %   n;   
  }   
    
  inline     
  long   impmod(long   restdig,   long   n)   
  {   
  long   d   =   restdig   /   DECS;   
  long   m   =   restdig   %   DECS;   
    
  long   mod   =   lpow(fpow[DECS]   %   n,   d,   n)   %   n;