hdu1059 Dividing

Posted 2020-10-05 ovo

Dividing

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

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.

 

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题解：

多重背包，把每个物品按数量的二进制拆成很多个物品进行dp，把价值和体积都设为价值的大小，即求价值不超过v(弹珠总价值的一半)时的最大价值，f[i]表示价值小于等于i时的最大价值

如果f[v]==v，则可行。

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
using namespace std;
int f[200000],num[10],w[280000],tot=0,v;
int get()
{
	int ans=0,f=1;char ch=getchar();
	while(!isdigit(ch)) {if(ch==‘-‘) f=-1;ch=getchar();}
	while(isdigit(ch)) {ans=ans*10+ch-‘0‘;ch=getchar();}
	return ans*f;
}
void init()
{
	int wi=0,zc;
	for(int i=1;i<=6;i++)
	if(num[i])       //////num[i]可能为0 
	{
		wi=0;           /////记得清0
		w[++tot]=i;
		zc=num[i]-1;        //减去的1是第一位的1 
		while(num[i])
		{
			num[i]>>=1;
			wi++;
			if(num[i]>1)
		    {
		    	w[++tot]=(1<<wi)*i;
		    	zc-=(1<<wi);
			}
			else if(num[i]==1)
			  w[++tot]=zc*i;
		}
	}
	fill(f,f+v+1,0);
}
int main()
{
	bool flag;
	for(int k=1;;k++)
	{
		flag=0;v=0;tot=0;
		for(int i=1;i<=6;i++)
		{
			num[i]=get();
			if(num[i])
			  flag=1;
			v+=num[i]*i;
		}
		if(!flag) return 0;
		printf("Collection #%d:\n",k);
		if(v%2!=0)
		{
			printf("Can‘t be divided.\n\n");
			continue;
		}
		else v>>=1;
		init();
		for(int i=1;i<=tot;i++)
		  for(int j=v;j>=0;j--)
		    if(w[i]<=j)
		      f[j]=max(f[j],f[j-w[i]]+w[i]);
		if(f[v]==v)
		  printf("Can be divided.\n\n");
		else
		  printf("Can‘t be divided.\n\n");
	}
	return 0;
}
/*
20001 10 1 1000 100 1 
*/