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): 18296    Accepted Submission(s): 5100




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.

把题目的value 当作 cost。

初始化 dp[0] =1 。

然后模版一套就好了。取v/2为容量 ，不然会超时。

#include<stdio.h>
#include<string.h>
#define N 10 //件数
#define NN 30000*6  //件数乘价格乘数量
int dp[NN],i,k,v;
void bag01(int cost)
{
	for(i=v;i>=cost;i--)
		dp[i]+=dp[i-cost];
}
void complete(int cost)
{
	for(i=cost;i<=v;i++) 
		dp[i]+=dp[i-cost];
}
void multiply(int cost,int amount)
{
	if(cost*amount>=v)
		complete(cost);
	else{
		k=1;
		while(k<amount){  //k分成1 2 4 8 16 个  二进制数,可以代表所有情况. 
			bag01(k*cost);
			amount-=k;
			k+=k;
		}
		if(amount!=0)
		bag01(cost*amount);
	}
}
int main()
{
	int n,w[N],num[N],cost[N],j;
	int cas=1;
	int sum;
	while(scanf("%d",&num[1])!=EOF){
		memset(dp,0,sizeof(dp));
		dp[0]=1;
		sum=0;
		sum+=num[1];
		cost[1]=1;
		for(j=2;j<=6;j++){
			scanf("%d",&num[j]);
			cost[j]=j;
			sum+=num[j]*j;
		}
		if(sum==0)
			break;

		v=sum/2;
		if(sum%2==0)
		for(j=1;j<=6;j++)
			multiply(cost[j],num[j]); 
		
		printf("Collection #%d:\n",cas++);
		if(dp[v]==0||sum%2==1)
			printf("Can't be divided.\n");
		else 
			printf("Can be divided.\n");
		puts("");
	}
	return 0;
}