The Balance （hdu1709）母函数

最新推荐文章于 2020-09-28 07:21:36 发布

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本文介绍了一种使用多个砝码进行精确药物剂量测量的方法。通过动态规划算法确定无法测量的质量范围，并提供具体实现代码示例。

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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


给出n个砝码，然后问你1-sum那些质量称不出来，

今天就是在刷母函数，复习复习 哈哈
这个我们要多考虑一点  就是砝码可以左右放，也就是多了 tmpp=fabs(j-k*s[i]);tmp[tmpp]+=ans[j];  然后把减去这种情况想到就好了

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
int s[10005],ans[10005],tmp[10005],cmax;
int num[10005],n;
int mu()
{
    memset(ans,0,sizeof(ans));
    memset(tmp,0,sizeof(tmp));
    ans[0]=1;
    cmax=0;
    for(int i=1; i<=n; i++)
    {
        cmax+=s[i];
        for(int j=0; j<=cmax; j++)
            for(int k=0; k<=1&&k*s[i]+j<=cmax; k++)
            {
                int tmpp=fabs(j-k*s[i]);
                tmp[j+k*s[i]]+=ans[j];
                tmp[tmpp]+=ans[j];//注意这里
            }
        memcpy(ans,tmp,sizeof(tmp));
        memset(tmp,0,sizeof(tmp));
    }
    return 0;
}
int main()
{
    while(cin>>n)
    {
        if(n==0)
            break;
        int sum=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&s[i]);
            sum+=s[i];
        }
        mu();
        int ssum=0;
        for(int i=1;i<=sum;i++)
        {
            if(ans[i]==0)
            ssum++;
        }
        cout<<ssum<<endl;
        int k=1;
        if(ssum!=0)
        {
            for(int i=1;i<=sum;i++)
            {
                 if(ans[i]==0&&k<ssum)
                 {
                      cout<<i<<" ";
                      k++;
                 }
                 else if(ans[i]==0)
                 {
                     cout<<i<<endl;
                     break;
                 }
            }
        }
    }
    return 0;
}