hdu1709 The Balance （放入+取出 01背包）

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 6973    Accepted Submission(s): 2874

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

 

Source

HDU 2007-Spring Programming Contest

解析：1.正向：往背包里面放东西，求出那些经过相加之后能得到的体积。

           2.逆向：从背包里面拿出东西，求出哪些组合相减之后能得到的体积。

代码：

#include<cstdio>#include<queue>#include<cstring>#define ms(a) memset(a,0,sizeof(a))using namespace std;const int maxn=1e2;const int maxsum=1e4;int a[maxn+10];bool f[maxsum+10];queue<int> q;int main(){  freopen("1.in","r",stdin);  int n,i,j,sum;  while(scanf("%d",&n)==1)    {      for(sum=0,i=1;i<=n;i++)    scanf("%d",&a[i]),sum+=a[i];  ms(f),f[0]=1;      for(i=1;i<=n;i++)        for(j=sum;j>=a[i];j--)f[j]|=f[j-a[i]];      for(i=1;i<=n;i++)        for(j=1;j+a[i]<=sum;j++)f[j]|=f[j+a[i]];              for(j=0,i=1;i<=sum;i++)if(!f[i])q.push(i),j++;      printf("%d\n",j);      if(j==0)continue;      printf("%d",q.front()),q.pop();      while(!q.empty())printf(" %d",q.front()),q.pop();      printf("\n");}  return 0;}