poj1014(多重背包问题）

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, outputCollection #k:’’, where k is the number of the test case, and then either Can be divided.'' orCan’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.

分析：可转化为一半的体积最多可以放进多少石头，如果最多刚好可以装满，则可以平分，反则不行。多重背包问题，动态分析，设一个数组p[i]存有i体积最多可以存多少体积的石头，然后将石头逐一考虑在i体积下要不要装它，但需要注意 的时，可能石头非常多，会导致超时，好在有重复的，放一个石头后，再分别放两个跟它相同体积的石头和把这两个石头捆绑一起放的石头效果是一样的，捆绑四个的时候也是如此，以此类推。

ac代码：

#include<iostream>
	#include<cstdio>
	#include<cstring>
	#include<algorithm>
	#include<queue>
	#include<string>
	using namespace std;
	int a[10], b[10], sum,half;
	int p[50000];
	void asd(int a)
	{
		for (int i = half; i >=a; i--)
			p[i] = (p[i - a] + a) > p[i] ? (p[i - a] + a) : p[i];
		
	}
	void ask(int a,int c)
	{
		int x=1;
		asd(a);
		c = c - 1;
		while (x<c)
		{
			x = 2 * x;
			asd(x*a);
			c = c - x;
		
		}if(c>0)
		asd(c*a);
	}

	int main()
	{
		int i = 0;
		while (cin >> b[0] >> b[1] >> b[2] >> b[3] >> b[4] >> b[5])
		{
			if (b[0] == 0 && b[1] == 0 && b[2] == 0 && b[3] == 0 && b[4] == 0 && b[5] == 0)
				break;
			sum = 0;
			memset(p, 0, sizeof(p));
			i++;
			for (int i = 1; i <= 6; i++)
			{
				a[i - 1] = i;
				sum += a[i - 1] * b[i - 1];
			}
			if(sum%2)
			cout << "Collection #" << i << ":" << endl<<"Can't be divided."<<endl<<endl;
			else {
				half = sum / 2;
				for (int i = 0; i < 6; i++){
					if (b[i] > 0)ask(i + 1, b[i]);
			}
				if (p[half] == half)cout << "Collection #" << i << ":" << endl << "Can be divided." << endl<<endl;
				else cout << "Collection #" << i << ":" << endl << "Can't be divided." << endl<<endl;
			}
		}
	}