The Balance

比眉伴天荒 2022-08-07 14:35 150阅读 0赞

# The Balance #

**Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 6235    Accepted Submission(s): 2564**

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

母函数，但是这一点要讲一下！不同之处，这里又多了一个减法

（除法），所以无从下手，一开始用深搜，数据太大，之间崩溃

了！

#include<cstdio>
    #include<cmath>
    #include<iostream>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    int n,sum,b[100005],a[1005],j,flag,sum1,c[100005];
    bool visit[1005];
    int compare(int a,int b)
    {
        return a>b;
    }
    void dfs(int pos)
    {
        int i;
        for(i=pos;i<n;i++)
        {
            sum1+=a[i];
            if(sum1==a[0])
            {
                flag++;
            }
            if(flag==2)
            {
                return;
            }
            b[j++]=sum1;
            visit[i]=1;
            dfs(i+1);
            sum1-=a[i];
            visit[i]=0;
        }
    }
    int main()
    {
        int i,k,x,y,flag2,sum2,sum3;
        while(cin>>n)
        {
            memset(a,0,sizeof(a));
            memset(b,0,sizeof(b));
            memset(c,0,sizeof(c));
            memset(visit,0,sizeof(visit));
            x=flag=j=sum1=sum=0;
            for(i=0;i<n;i++)
            {
                cin>>a[i];
                sum+=a[i];
            }
            sort(a,a+n,compare);
            dfs(0);
          /*  for(i=0;i<n;i++)
            {
                sum2=0;
                for(k=i+1;k<n;k++)
                {
                    sum3=a[k];
                    sum2+=sum3;
                    if(sum2<a[i])
                    {
                        b[j++]=a[i]-sum2;
                    }
                    if(sum3<a[i]&&sum3!=sum2)
                    {
                        b[j++]=a[i]-sum3;
                    }
                }
            }*/
            for(i=1;i<=sum;i++)
            {
                y=0;
                for(k=0;k<j;k++)
                {
                    if(i==b[k])
                    {
                        y=1;
                        break;
                    }
                }
                if(y==0)
                {
                    c[x++]=i;
                }
            }
            if(x==0)
            {
                cout<<0<<endl;
            }
            else
            {
                cout<<x<<endl;
                for(i=0;i<x-1;i++)
                {
                    cout<<c[i]<<" ";
                }
                cout<<c[x-1]<<endl;
            }
        }
    }

看到网上的代码，感觉写的很好！也深深地体会到了母函数

#include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int a[105],b[10005],c[10005],ans[10005];
    int main()
    {
        int i,j,f,n,k,sum;
        while(cin>>n)
        {
            f=sum=0;
            for(i=1;i<=n;i++)
            {
                cin>>a[i];
                sum+=a[i];
            }
            memset(b,0,sizeof(b));
            memset(c,0,sizeof(c));
            b[0]=1;b[a[1]]=1;//初始化
            for(i=2;i<=n;i++)
            {
                for(j=0;j<=sum;j++)
                {
                   for(k=0;k+j<=sum&&k<=a[i];k+=a[i])//这里之所以限制,所以设置k<=a[i]实际上是为了避免多次加a[i]
                   {
                       c[k+j]+=b[j];
                       c[abs(j-k)]+=b[j];
                   }
                }
                for(j=0;j<=sum;j++)
                {
                    b[j]=c[j];
                    c[j]=0;
                }
            }
            for(i=1;i<=sum;i++)
            {
                if(!b[i])
                {
                    ans[f++]=i;
                }
            }
            if(f==0)
            {
                cout<<0<<endl;
            }
            else
            {
                cout<<f<<endl;
                for(i=0;i<f-1;i++)
                {
                    cout<<ans[i]<<" ";
                }
                cout<<ans[f-1]<<endl;
            }
        }
    }