hdu 1059 Dividing

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.

 

题解：

   第一次写多重背包的问题，其中要考虑是否能被均分，用到物品的价值和所占体积相等，使得判断更加简单。不多说，上代码。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
int dp[120000],V;
void bag01(int v,int w)                                                //01背包
{
    for(int i=V; i>=v; i--)
        dp[i]=max(dp[i],dp[i-v]+w);
}
void bag11(int v,int w)                                              //完全背包
{
    for(int i=v; i<=V; i++)
        dp[i]=max(dp[i],dp[i-v]+w);
}
void bagmuti(int m,int v,int w)                               //多重背包
{
    if(m*v>=V)
        bag11(v,w);
    else
    {
        int k=1;
        while(k<m)
        {
            bag01(k*v,k*w);
            m-=k;
            k*=2;
        }
        bag01(m*v,m*w);
    }

}


int main()
{
    int a[7],t=1;
    while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6]),a[1]+a[2]+a[3]+a[4]+a[5]+a[6])
    {
        int sum=a[1]+a[2]*2+a[3]*3+a[4]*4+a[5]*5+a[6]*6;
        if(sum%2)                                                                                           //总价值为奇数，直接判断输出
        {
            printf("Collection #%d:
Can't be divided.

",t++);
            continue;
        }
        V=sum/2;
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=6; i++)
            bagmuti(a[i],i,i);                                                                               //体积和价值是一样的
        if(dp[V]==V)                                                                                       //此处为技巧，定义时物件的价值和所需体积是一样的，所以可以这样用
            printf("Collection #%d:
Can be divided.

",t++);
        else
            printf("Collection #%d:
Can't be divided.

",t++);

    }

    return 0;
}