POJ 1276 Cash Machine

POJ 1276 Cash Machine

多重背包问题。将ni拆分成1，2，4，...，2^(k-1)，ni-2^k+1，易证这种拆分可以组合出1..ni之间的所有数字。于是转化成01背包求解。
以下是我的代码：

/*
 * Author:  lee1r
 * Created Time:  2011/8/9 10:20:19
 * File Name: poj1276.cpp
  */
#include < iostream >
#include < sstream >
#include < fstream >
#include < vector >
#include < list >
#include < deque >
#include < queue >
#include < stack >
#include < map >
#include < set >
#include < bitset >
#include < algorithm >
#include < cstdio >
#include < cstdlib >
#include < cstring >
#include < cctype >
#include < cmath >
#include < ctime >
#define  L(x) ((x)<<1)
#define  R(x) (((x)<<1)+1)
#define  Half(x) ((x)>>1)
#define  lowbit(x) ((x)&(-(x)))
using   namespace  std;
const   int  kInf( 0x7f7f7f7f );
const   double  kEps(1e - 8 );
typedef  long   long  int64;
typedef unsigned  long   long  uint64;

const   int  kMaxn( 10007 );
const   int  kMaxm( 100007 );

int  n,m,cnt,w[kMaxn],c[kMaxn],d[kMaxm];

int  main()
{
     // freopen("data.in","r",stdin);
    
     while (scanf( " %d " , & m) == 1 )
    {
        cnt = 0 ;
        scanf( " %d " , & n);
         for ( int  i = 1 ;i <= n;i ++ )
        {
             int  ni,wi;
            scanf( " %d%d " , & ni, & wi);
             for ( int  i = 1 ;i <= ni;ni -= i,i <<= 1 )
            {
                cnt ++ ;
                w[cnt] = wi * i;
                c[cnt] = wi * i;
            }
             if (ni > 0 )
            {
                cnt ++ ;
                w[cnt] = wi * ni;
                c[cnt] = wi * ni;
            }
        }
        
         int  ans( 0 );
        memset(d, 0 , sizeof (d));
         for ( int  i = 1 ;i <= cnt;i ++ )
             for ( int  j = m;j >= c[i];j -- )
                d[j] = max(d[j],d[j - c[i]] + w[i]);
         for ( int  i = 0 ;i <= m;i ++ )
            ans = max(ans,d[i]);
        
        printf( " %d\n " ,ans);
    }
    
     return   0 ;
}