Dividing （hdu 1059 多重背包）

Dividing

Sample Input

1 0 1 2 0 0　　　价值为1，2，3，4，5，6的物品数目分别为 1 0 1 2 0 0，求能否将这些物品按价值分为两堆，转化为多重背包。
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.

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int dp[120005];
int a[7];
int value=0;
void completepack(int cost,int weight)        //代价cost换取价值weight 完全背包
{
    for(int i=cost;i<=value;i++)
        dp[i]=max(dp[i],dp[i-cost]+weight);
}
void zeroonepack(int cost,int weight)        
{
    for(int i=value;i>=cost;i--)
        dp[i]=max(dp[i],dp[i-cost]+weight);
}
void multiplepack(int cost,int weight,int amount)        
{
    int t=1;
    if(amount*cost>=value)        
        completepack(cost,weight);        
    else
    {
        while(t<amount)
        {
            zeroonepack(cost*t,cost*t);            
            amount-=t;
            t<<=1;
        }
        if(amount)
            zeroonepack(cost*amount,cost*amount);
    }
}
int main()
{
    int i,j,t=0;
    freopen("in.txt","r",stdin);
    while(scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6]))
    {
        int sum=0;
        for(i=1;i<=6;i++)
            sum+=a[i]*i;
        value=sum/2;
        if(sum==0)
            break;
        printf("Collection #%d:
",++t);
        if(sum%2)
        {
            printf("Can't be divided.

");
            continue;
        }
        memset(dp,0,sizeof(dp));
        for(i=1;i<=6;i++)
        {
            if(a[i])
                multiplepack(i,i,a[i]);
        }
        if(dp[value]==value)
            printf("Can be divided.

");
        else
            printf("Can't be divided.

");
    }
    return 0;
}