HDOJ 1709 The Balance(母函数)



The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7060    Accepted Submission(s): 2912

Problem Description

Now you are asked to measure a dose of medicine with a balance and a number of weights. Certainly it is not always achievable. So you should find out the qualities which cannot be measured from the range [1,S]. S is the total quality of all the weights.

 

Input

The input consists of multiple test cases, and each case begins with a single positive integer N (1<=N<=100) on a line by itself indicating the number of weights you have. Followed by N integers Ai (1<=i<=N), indicating the quality of each weight where 1<=Ai<=100.

 

Output

For each input set, you should first print a line specifying the number of qualities which cannot be measured. Then print another line which consists all the irrealizable qualities if the number is not zero.

 

Sample Input

   
   
   
   

    
    
    
    3 1 2 4 3 9 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0 2 4 5
   
   
   
   

题意： 有一个天平和n个砝码，给出这些砝码的质量ai，现在问你在[1，s]（s表示砝码质量之和）间有哪些质量是秤不出来的，先输出个数，在输出确定的数。

题解：尴尬，这么简单的题都没读懂。 母函数变形啦，砝码是放在天平的左右两边的，对于每种质量我们可以ai+aj==weight，也可以 ai-aj==weight。    所以在进行合并乘法式是注意  c2[k+j]+=c1[j]， 也有c2[abs(j-k)]+=c1[j]。

代码如下：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#define maxn 10010
int c1[maxn],c2[maxn];
int a[110];
int main()
{
	int n,i,j,k,cnt,sum,num;
	while(scanf("%d",&n)!=EOF)
	{
		sum=0;
		for(i=1;i<=n;++i)
		{
			scanf("%d",&a[i]);
			sum+=a[i];
		}
		memset(c1,0,sizeof(c1));
		memset(c2,0,sizeof(c2));
		c1[0]=c1[a[1]]=1;
		num=a[1];
		for(i=2;i<=n;++i)
		{
			for(j=0;j<=num;++j)
			{
				for(k=0;k<=a[i]&&k+j<=sum;k+=a[i])
				{
					c2[k+j]+=c1[j];
					c2[abs(j-k)]+=c1[j];
				}
			}
			num+=a[i];
			for(j=0;j<=num;++j)
			{
				c1[j]=c2[j];
				c2[j]=0;
			}
		}
		int cnt=0;
		for(i=0;i<=sum;++i)
		{
			if(!c1[i])
				c2[cnt++]=i; 
		}
		if(cnt==0)
			printf("0\n");
		else
		{
			printf("%d\n",cnt);
			for(i=0;i<cnt;++i)
			{
				if(i==cnt-1)
					printf("%d\n",c2[i]);
				else
					printf("%d ",c2[i]);
			}
		}
	}
	return 0;
}