母函数——秤砣问题——The Balance

最新推荐文章于 2021-08-01 13:55:03 发布

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最新推荐文章于 2021-08-01 13:55:03 发布

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分类专栏：母函数

本文链接：https://blog.youkuaiyun.com/codetypeman/article/details/81612214

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该问题探讨如何使用一组重量的秤砣测量不同质量的药物剂量。输入包含秤砣的数量及其重量，目标是找出在不超过总重量S的情况下无法测量的质量。输出是不可测量的质量数及其具体值。示例说明了如何通过秤砣组合来确定可测量的质量范围。

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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

题意：给你一些秤砣，找出在不超过重质量的情况下，哪些重量是称不出来的

考虑一个特殊的情况：给你一个1，一个2，要想称出1，很明显放一个重量为1的即可，但是天平左边放1，右边放2，同样可以称出1；

可以称出的质量还可得到这样一个式子 $M=\left | m1-m2 \right |$ ; m1,m2 分别为两个秤砣重量；

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=20000;
int n;
int a[maxn];
int t[maxn];
int v[maxn];

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int s=0;
        memset(t,0,sizeof t);
        memset(v,0,sizeof v);
        for(int i=1;i<=n;++i)
            scanf("%d",&a[i]),s+=a[i];
        v[0]=1;
        v[a[1]]=1;
        for(int i=2;i<=n;++i)
        {
            for(int j=0;j<=s;++j)
                for(int k=0;k+j<=s&&k<=a[i];k+=a[i])
            {
                t[abs(j-k)] += v[j];
                t[j+k]+=v[j];
            }
            for(int j=0;j<=s;++j)
            {
                v[j]=t[j];
                t[j]=0;
            }
        }
        int cnt=0;
        for(int i=0;i<=s;++i)
            if(v[i]==0) ++cnt;
        cout<<cnt<<endl;
        if(cnt==0) continue;
        else
        {
            for(int i=0;i<=s;++i)
            {
                if(!v[i])
                {
                    printf("%d",i);
                    --cnt;
                    if(cnt) printf(" ");
                    else puts("");
                }
            }
        }
    }
}