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

31 2 439 2 1

 

Sample Output

024 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];elseb[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！