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组合数学 - 母函数的运用 --- hdu 1709 :The Balance

2024-07-23 10:41:05   221人阅读

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5706    Accepted Submission(s): 2311


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
 

 

Mean: 

 给你N个砝码，每个砝码有自己的重量，现在要你计算从1~S中哪些重量是不能用这些砝码称出来的。(其中S为所有砝码的重量之和)。

analyse:

 就是一道母函数的运用题，要注意的是砝码可以摆放在两个托盘上，所以要注意的是两个托盘两边的差值也能称出来的情况，其他的和普通母函数差不多。

Time complexity：O(n^3)

 

Source code：

 

// Memory   Time// 1347K     0MS// by : Snarl_jsb// 2014-09-19-11.59#include<algorithm>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<vector>#include<queue>#include<stack>#include<map>#include<string>#include<climits>#include<cmath>#define N 10100#define LL long longusing namespace std;int val[N];int c1[N],c2[N];int main(){    ios_base::sync_with_stdio(false);    cin.tie(0);//    freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);//    freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);    int n;    while(cin>>n)    {        long long sum=0;        for(int i=1;i<=n;++i)        {            cin>>val[i];            sum+=val[i];        }        memset(c1,0,sizeof(c1));        memset(c2,0,sizeof(c2));        c1[0]=c1[val[1]]=1;        for(int i=2;i<=n;++i)        {            for(int j=0;j<=sum;++j)            {                for(int k=0;k<=1;++k)                {                    c2[k*val[i]+j]+=c1[j];                    c2[abs(k*val[i]-j)]+=c1[j];                }            }            for(int j=0;j<=sum;++j)            {                c1[j]=c2[j];                c2[j]=0;            }        }        int res[N],ans=0;        for(int i=1;i<=sum;++i)        {            if(!c1[i])            {                res[ans++]=i;            }        }        if(!ans)        {            puts("0");            continue;        }        cout<<ans<<endl;        ans--;        for(int i=0;i<ans;++i)        {            cout<<res[i]<<" ";        }        cout<<res[ans]<<endl;    }    return 0;}

　　

 

组合数学 - 母函数的运用 --- hdu 1709 :The Balance


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