Dividing (经典多重背包，处理大数据，模余最小公倍数)

1、http://acm.hdu.edu.cn/showproblem.php?pid=1059

处理大数据，模余最小公倍数

2、题目大意：

Marsha 和Bill收集了很多 弹球，现在他们想把这些弹球分开，要求两个人分的一样多，如果所有的弹球都有相同的价值，那么分将很容易，一人得到一半即可，但是不幸的是，有些弹球很大，有些比别的漂亮，现在他们给每个弹球都估算了一个价值，价值分别是1、2、3、4、5、6，现在他们想把这些弹球分成两部分，要求两人的价值相同，同时他们意识到分开很难，因为如果是奇数将分不开，即便是偶数，也不一定能分开，例如，有一个价值为1的，一个价值为3的，两个价值为4的，就不能分成相等的两部分，因此需要你写一个程序判断能不能将这些弹球分成价值相等的两部分

 

输入：每行包括6个非负数的整数，n1, n2, ..., n6，ni代表价值为i的弹球有ni个，个数可能最大达到20000

 

输出：
每个样例输出``Collection #k:''，k是第几个样例数，然后输出``Can be divided.'' or ``Can't be divided.''.
3、题目：

Dividing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10834    Accepted Submission(s): 3035

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.
   
   
   
   

 

4、代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define minn -9999999
using namespace std;
int v[120005];
int dp[420005];
int main()
{
    int  a[10],cas=0;
    while(1)
    {
        cas++;
        int flag=0,sum=0;
        memset(v,0,sizeof(v));

        for(int i=1; i<=6; i++)
        {
            scanf("%d",&a[i]);
            a[i]=a[i]%60;//%最小公倍数
            sum+=a[i]*i;
        }
        for(int i=1; i<=sum/2+1; i++)
            dp[i]=minn;
        dp[0]=0;
        if(sum%2!=0)
            flag=1;
        if(a[1]==0&&a[2]==0&&a[3]==0&&a[4]==0&&a[5]==0&&a[6]==0)
            break;
        int k=1;
        for(int i=1; i<=6; i++)
        {
            for(int j=1; j<=a[i]; j=j*2)
            {
                v[k++]=i*j;
                a[i]-=j;
            }
            if(a[i]>0)
            {
                v[k++]=a[i]*i;
            }
        }
        for(int i=1; i<=k; i++)
        {
            for(int j=sum/2; j>=0; j--)
            {
                if(j>=v[i])
                    dp[j]=max(dp[j],dp[j-v[i]]+v[i]);
            }
        }
        printf("Collection #%d:\n",cas);
        if(dp[sum/2]<0||flag==1)
        {
            printf("Can't be divided.\n");
            printf("\n");
        }
        else
        {
            printf("Can be divided.\n");
            printf("\n");
        }
    }
    return 0;
}
/*
1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0
*/