POJ 1276 Cash Machine

解题思路:dp(多重背包问题)

将多重背包转化为01背包，假设每个物品的数量上限为n，每件的体积为0，价值为w

那么可以把该物品分为1,2,4,...,2^(k-1)和n-(2^k+1),那么新的物品就是(体积，价值):(0,w),(0,2w),...,(0,2^(k-1)w)和(0,(n-2^k+1)w)

(这样可以构造出任何<=n的情况，假设n为7，则分成系数为1,2和4,那么1-7的所有情况均可由1,2,4这3个基数组合得到)

NULL

   
     

    
      
    
      #include 
    
      <
    
      iostream
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;

    
      #define
    
       MAXN 100001
    
      

    
      int
    
       main()
{
 
    
      int
    
       c,m,i,j,k,t,p;
 
    
      int
    
       d[
    
      11
    
      ],n[
    
      11
    
      ],dp[MAXN];
 
    
      while
    
      (scanf(
    
      "
    
      %d %d
    
      "
    
      , 
    
      &
    
      c, 
    
      &
    
      m)
    
      !=
    
      EOF)
 {
 
    
      for
    
      (i
    
      =
    
      0
    
      ;i
    
      <
    
      m;i
    
      ++
    
      )scanf(
    
      "
    
      %d %d
    
      "
    
      , 
    
      &
    
      n[i], 
    
      &
    
      d[i]);
 memset(dp, 
    
      0
    
      , 
    
      sizeof
    
      (dp));
 
    
      for
    
      (i
    
      =
    
      0
    
      ;i
    
      <
    
      m;i
    
      ++
    
      )
 
    
      if
    
      (n[i]
    
      *
    
      d[i]
    
      ==
    
      c){dp[c]
    
      =
    
      c;
    
      break
    
      ;}
 
    
      else
    
       
    
      if
    
      (n[i]
    
      *
    
      d[i]
    
      >
    
      c)
 {
 
    
      for
    
       (j
    
      =
    
      d[i];j
    
      <=
    
      c;j
    
      ++
    
      )
 
    
      if
    
      (dp[j
    
      -
    
      d[i]]
    
      +
    
      d[i]
    
      >
    
      dp[j])dp[j]
    
      =
    
      dp[j
    
      -
    
      d[i]]
    
      +
    
      d[i];
 }
 
    
      else
    
      
 {
 
    
      for
    
      (t
    
      =
    
      n[i],j
    
      =
    
      1
    
      ;j
    
      <=
    
      (n[i]
    
      >>
    
      1
    
      );t
    
      -=
    
      j,j
    
      *=
    
      2
    
      )
 
    
      for
    
      (k
    
      =
    
      c,p
    
      =
    
      j
    
      *
    
      d[i];k
    
      >=
    
      p;k
    
      --
    
      )
 
    
      if
    
      (dp[k
    
      -
    
      p]
    
      +
    
      p
    
      >
    
      dp[k])dp[k]
    
      =
    
      dp[k
    
      -
    
      p]
    
      +
    
      p;
 
    
      for
    
      (k
    
      =
    
      c,p
    
      =
    
      t
    
      *
    
      d[i];k
    
      >=
    
      p;k
    
      --
    
      )
 
    
      if
    
      (dp[k
    
      -
    
      p]
    
      +
    
      p
    
      >
    
      dp[k])dp[k]
    
      =
    
      dp[k
    
      -
    
      p]
    
      +
    
      p;
 }
 printf(
    
      "
    
      %d\n
    
      "
    
      ,dp[c]);
 }
 
    
      return
    
       
    
      0
    
      ;
}