hdu 1059 Dividing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19391    Accepted Submission(s): 5446

Problem Description

Marsha
and Bill own a collection of marbles. They want to split the collection
among themselves so that both receive an equal share of the marbles.
This would be easy if all the marbles had the same value, because then
they could just split the collection in half. But unfortunately, some of
the marbles are larger, or more beautiful than others. So, Marsha and
Bill start by assigning a value, a natural number between one and six,
to each marble. Now they want to divide the marbles so that each of them
gets the same total value.
Unfortunately, they realize that it
might be impossible to divide the marbles in this way (even if the total
value of all marbles is even). For example, if there are one marble of
value 1, one of value 3 and two of value 4, then they cannot be split
into sets of equal value. So, they ask you to write a program that
checks whether there is a fair partition of the marbles.

 

Input

Each
line in the input describes one collection of marbles to be divided.
The lines consist of six non-negative integers n1, n2, ..., n6, where ni
is the number of marbles of value i. So, the example from above would
be described by the input-line ``1 0 1 2 0 0''. The maximum total number
of marbles will be 20000.

The last line of the input file will be ``0 0 0 0 0 0''; do not process this line.

 

Output

For
each colletcion, output ``Collection #k:'', where k is the number of
the test case, and then either ``Can be divided.'' or ``Can't be
divided.''.

Output a blank line after each test case.

 

Sample Input

1 0 1 2 0 0

1 0 0 0 1 1

0 0 0 0 0 0

 

Sample Output

Collection #1:

Can't be divided.

Collection #2:
Can be divided.

 

Source

Mid-Central European Regional Contest 1999

 思路：之前还没看多重背包，就直接把每种物品分为 （背包大小/费用）件,然，我们并不用把每种物品分为费用和权值都相同的多件，为尽量减少分的件数，可每件的费用为 c[i]×2^k, 价值为 w[i]×2^k .......n[i] - 2^k + 1, 如n[i] = 13 时， 可分为 1 ， 2，  4，  6  作为系数分别乘以 c[i],  w[i]

 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <iostream>
 #include <cmath>
 #include <algorithm>
 using namespace std ;
 int dp[] ;
 int r[] ;
 int sum, num ;
 int c =  ;
 void _zeroone(int c, int w)//0-1背包处理单件物品
 {
     for(int i = sum; i >= c; --i)
         dp[i] = max(dp[i], dp[i - c] + w) ;
 }
 void _compl(int c, int w)//完全背包处理单件物品
 {
     for(int i = c; i <= sum; ++i)
         dp[i] = max(dp[i], dp[i - c] + w) ;
 }
 void  _mult(int c, int w, int amount)
 {
     if(c * amount >= sum) { _compl(c, w) ; return ; }//满足条件时,该件物品可以直接当完全背包时的单件物品处理
     int k =  ;
     while(k < amount)//二进制分解
     {
         _zeroone(c * k,w * k) ;
         amount -= k ;
         k = k *  ;
     }
     _zeroone(amount * c, amount * w) ;
 }
 int main()
 {
     while()
     {
         memset(dp, , sizeof dp) ;
         sum =  ;
         int flag =  ;
         for(int i = ; i <= ; ++i){
             scanf("%d",&r[i]) ;
             if(r[i] != ) flag =  ;
             sum += i * r[i] ;
         }
         if(!flag) break ;
         printf("Collection #%d:\n",c++) ;
         if(sum & ) { printf("Can't be divided.\n\n") ; continue ;}
         sum /=  ;
         for(int i = ; i <= ; ++i)//将多重背包每件分开转换为完全背包或0-1背包中的单件物品处理
         _mult(i,i,r[i]) ;
         if(dp[sum] == sum ) printf("Can be divided.\n") ;
         else printf("Can't be divided.\n") ;
         printf("\n") ;
     }
     return  ;
 }