HDU 1709 The Balance

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

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string.h>
using namespace std;
int main()
{
    int a[10003],r[10003],b[103];//开始WA了，检查了N久，试了好多数据都没问题
    int i,k,n,sum;                           //对比了下别人的代码，发现我的r[]写错了10003写成1003，囧
    while(scanf("%d",&n)!=EOF)
    {    sum=0;
        for(i=1;i<=n;sum+=b[i],i++)
          scanf("%d",&b[i]);
        memset(a,0,sizeof(a));
        memset(r,0,sizeof(r));
       r[0]=a[0]=1;
       r[b[1]]=a[b[1]]=1;
        for(k=2;k<=n;k++)
         {
             for(i=0;i+b[k]<=sum;i++)
               {
                   a[i+b[k]]+=r[i];
                   a[int(fabs(1.0*i-b[k]))]+=r[i];//关键点
                }
                for(i=0;i<=sum;i++)
                   r[i]=a[i];
         }
        for(k=0,i=1;i<=sum;i++)
          if(!a[i])
           {
               r[k++]=i;
           }
        printf("%d\n",k);
        if(k)
        {
          for(i=0;i<k-1;i++)
            printf("%d ",r[i]);
            printf("%d\n",r[i]);
        }
    }                    //这题这样做速度挺慢，差点超时， 再去想想其他办法或者优化下。
    return 0;
}

//刚刚去想了下，做出如下改进，快了近一半，而且少开了一个数组

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string.h>
using namespace std;
int main()
{
    freopen("in.txt","r",stdin);
    int a[10003],r[1003],b;
    int i,k,n,sum;
    while(scanf("%d",&n)!=EOF)
    {    sum=0;
    scanf("%d",&b); sum+=b;
    memset(a,0,sizeof(a));
    memset(r,0,sizeof(r));
    r[0]=a[0]=1;
    r[b]=a[b]=1;
        for(k=2;k<=n;k++)
          { scanf("%d",&b);
             sum+=b;
             for(i=0;i+b<=sum;i++)
               {
                   a[i+b]+=r[i];
                   a[int(fabs(1.0*i-b))]+=r[i];
                }
            for(i=0;i<=sum;i++)
                r[i]=a[i];
         }
        for(k=0,i=1;i<=sum;i++)
          if(!a[i])
           {
               r[k++]=i;
           }
        printf("%d\n",k);
        if(k)
        {
          for(i=0;i<k-1;i++)
            printf("%d ",r[i]);
            printf("%d\n",r[i]);
        }
    }
 return 0;
}