hdu1709(母函数)

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4751    Accepted Submission(s): 1900

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
   
   
   
   

 

Source

HDU 2007-Spring Programming Contest

 

Recommend

lcy

 

本题给出一个带有若干个的砝码的天平，问[1,sum]总共有多少个质量不能称量。

本题是个组合数问题，问这些给定的数能组合出哪些数，当然本题背景决定可以通过减来组合。这是本题的亮点。

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

int num[100+10];
int n,sum;
int a[100*100+100];
int b[100*100+100];

inline void Solve()
{
	int i,j,k;
	memset(a,0,sizeof(a));
	memset(b,0,sizeof(b));
	a[0]=1;
	for(i=1;i<=n;i++)///枚举单个药剂质量
	{
		for(j=0;j<=sum;j++)//枚举被乘数的所有质量
		{
			for(k=0;k*num[i]+j<=sum&&k<=1;k++)
			{
				if(k*num[i]>j)b[k*num[i]-j]+=a[j];//
				else b[j-k*num[i]]+=a[j];//想减的情况意味着放两边
				b[k*num[i]+j]+=a[j];//相加的情况意味着放一边
			}
		}
		memcpy(a,b,sizeof(b));
		memset(b,0,sizeof(b));
	}
}

int main()
{
	int i,cnt,k;
	while(~scanf("%d",&n))
	{
		sum=0;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&num[i]);
			sum+=num[i];
		}
		Solve();
		cnt=0;
		for(i=1;i<=sum;i++)
		{
			if(!a[i])
			{
				cnt++;
			}
		}
		printf("%d\n",cnt);
		k=0;
		for(i=1;i<=sum;i++)
		{
			if(!a[i]&&++k<cnt)
				printf("%d ",i);
			else if(!a[i]&&k==cnt)
			printf("%d\n",i);
		}
	}
	return 0;
}