HDU 1709 The Balance(母函数)

题目：HDU-1085

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1085

题目：

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7089    Accepted Submission(s): 2923

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
   
   
   
   

 

题目的意思呢，说实话看了很久才看懂，英语差就是心塞。意思是现在有n个砝码，每个质量为a[i]，问1~砝码质量之和 不能称出来的重量有几个，有的话输出来。

还是母函数，不过注意因为是天平称东西，所以不仅仅是加，减也是可以得哦~4kg和5kg的砝码也能称出来1kg的物品~~~

看代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<cstdio>
using namespace std;
const int maxn= 10005;
int a[maxn],b[maxn],ans[maxn];
int n,v,t,num;
int main(){
	while(cin>>n){
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		cin>>v;
		a[0]=1;
		a[v]=1;
		t=v;
		for(int i=1;i<n;i++){
			cin>>v;
			t+=v;
			for(int j=0;j<=t;j++){
				b[j]+=a[j];
				b[v+j]+=a[j];
				if(j-v>=0)                        //注意这里！
					b[j-v]+=a[j];
				else
					b[v-j]+=a[j];
			}
			for(int j=0;j<=t;j++){
				a[j]=b[j];
				b[j]=0;
			}
		}
		num=0;
		for(int i=0;i<t;i++)                            //输出别输错了按要求来
			if(a[i]==0){
				ans[num]=i;
				num++;
			}
		cout<<num<<endl;
		if(num!=0){
			for(int i=0;i<num-1;i++)
				cout<<ans[i]<<" ";
			cout<<ans[num-1]<<endl;
		}
	}
	return 0;
}

天天向上~~~~加油！