hdu 1059 Dividing（多重背包）

Dividing

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 19539 Accepted Submission(s): 5491



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.

有价值为1,2,3,4,5,6的六种物品 各有ni件 现在要将这些物品分为两个部分

每个部分的价值和相同

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 120010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int x = 0;
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}

int dp[MAXN];
int val[6]={1,2,3,4,5,6};
int n[6];
int m;

void ZeroOnePack(int wei,int val)
{
    for(int i=m;i>=wei;i--)
    {
        dp[i]=max(dp[i],dp[i-wei]+val);
    }
}

void CompletePack(int wei,int val)
{
    for(int i=wei;i<=m;i++)
    {
        dp[i]=max(dp[i],dp[i-wei]+val);
    }
}

void MultiPack(int wei,int val,int cnt)
{
    if(m<=wei*cnt)//总容量比物品所有的容量小 物品可以直到取完
    {
        CompletePack(wei,val);
        return ;
    }
    else
    {
        int k=1;
        while(k<=cnt)
        {
            ZeroOnePack(k*wei,k*val);
            cnt=cnt-k;
            k=2*k;
        }
        ZeroOnePack(cnt*wei,cnt*val);
    }
}

int main()
{
//    fread;
    int cs=1;
    while(scanf("%d",&n[0])!=EOF)
    {
        int sum=1*n[0];
        for(int i=1;i<6;i++)
        {
            scanf("%d",&n[i]);
            sum+=(i+1)*n[i];
        }
        int flag=0;
        for(int i=0;i<6;i++)
        {
            if(n[i])
            {
                flag=1;
                break;
            }
        }
        if(!flag) break;
        printf("Collection #%d:\n",cs++);
        if(sum%2)
        {
            puts("Can't be divided.");
            puts("");
            continue;
        }
//        cout<<"sum "<<sum<<endl;
        MEM(dp,0);
        m=sum;
        for(int i=0;i<6;i++)
        {
            MultiPack(i+1,i+1,n[i]);
        }
        sum/=2;
        if(dp[sum]==sum)
            puts("Can be divided.");
        else puts("Can't be divided.");
        puts("");
    }
    return 0;
}