The Balance

最新推荐文章于 2021-08-01 13:55:03 发布

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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

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;
        }
    }
}