hdoj 1709 The Balance <母函数变形--有点像找同因数--队列排数>

最新推荐文章于 2019-04-25 16:05:33 发布

leibniz_zhang

最新推荐文章于 2019-04-25 16:05:33 发布

阅读量344

点赞数

CC 4.0 BY-SA版权

分类专栏：栈和队列母函数杭电oj 打表

本文链接：https://blog.youkuaiyun.com/leibniz_zhang/article/details/51996151

杭电oj 同时被 3 个专栏收录

83 篇文章

订阅专栏

打表

20 篇文章

订阅专栏

栈和队列

15 篇文章

订阅专栏

本文介绍了一个算法问题，即如何使用平衡秤和一组砝码来测量药品剂量。文章提供了完整的源代码实现，并通过样例输入输出展示了算法的工作原理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

The Balance

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7476 Accepted Submission(s): 3103

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

Source

HDU 2007-Spring Programming Contest

打表筛数---

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MA 10100
int m[MA],kp;
int shu[MA];
int main()
{
    int n;
    while (~scanf("%d",&n))
    {
        int s=0;
        for (int i=0;i<n;i++)
        {
            scanf("%d",&shu[i]);
            s+=shu[i];
        }
        kp=0;
        memset(m,0,sizeof(m));
        int qu[MA];
        qu[kp++]=0;m[0]=1;
        int a,b,c;
        for (int i=0;i<n;i++)
        {
            int ll=kp;
            for (int j=0;j<ll;j++)
            {
                if (!m[qu[j]+shu[i]])
                {
                    qu[kp++]=qu[j]+shu[i];
                    m[qu[kp-1]]=1;
                }
                a=min(qu[j],shu[i]);
                b=max(qu[j],shu[i]);
                c=b-a;
                if (!m[c])
                {
                    qu[kp++]=c;
                    m[c]=1;
                }
            }
        }
        int ss=0;
        for (int i=1;i<=s;i++)
            if (!m[i])
            ss++;
        printf("%d\n",ss);
        for (int i=1;i<=s;i++)
        {
            if (!m[i])
            {
                ss--;
                if (ss)
                    printf("%d ",i);
                else
                    printf("%d\n",i);
            }
        }
    }
    return 0;
}