zoukankan      html  css  js  c++  java
  • zoj3706 Break Standard Weight

    Break Standard Weight

    Time Limit: 2 Seconds      Memory Limit: 65536 KB

    The balance was the first mass measuring instrument invented. In its traditional form, it consists of a pivoted horizontal lever of equal length arms, called the beam, with a weighing pan, also called scale, suspended from each arm (which is the origin of the originally plural term "scales" for a weighing instrument). The unknown mass is placed in one pan, and standard masses are added to this or the other pan until the beam is as close to equilibrium as possible. The standard weights used with balances are usually labeled in mass units, which are positive integers.

    With some standard weights, we can measure several special masses object exactly, whose weight are also positive integers in mass units. For example, with two standard weights 1 and 5, we can measure the object with mass 1, 4, 5 or 6 exactly.

    In the beginning of this problem, there are 2 standard weights, which masses are x and y. You have to choose a standard weight to break it into 2 parts, whose weights are also positive integers in mass units. We assume that there is no mass lost. For example, the origin standard weights are 4 and 9, if you break the second one into 4 and 5, you could measure 7 special masses, which are 1, 3, 4, 5, 8, 9, 13. While if you break the first one into 1 and 3, you could measure 13 special masses, which are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13! Your task is to find out the maximum number of possible special masses.

    Input

    There are multiple test cases. The first line of input is an integer T < 500 indicating the number of test cases. Each test case contains 2 integers x and y. 2 ≤ x, y ≤ 100

    Output

    For each test case, output the maximum number of possible special masses.

    //枚举两种拆分,各自dfs一次,hash记录一下

    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int path[4];
    bool vis[2001];
    int hash[2001];  

    void dfs(int sum)
    {
        hash[sum]=1;
        for(int i=1;i<=3;i++)
        {
            if(!vis[i])
            {
                vis[i]=1;
                dfs(sum+path[i]);
                vis[i]=0;
            }
            if(!vis[i]&&(sum-path[i])>0)
            {
                vis[i]=1;
                dfs(sum-path[i]);
                vis[i]=0;
            }
        }
    }

    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
            int n,m;
             scanf("%d%d",&n,&m);
            int mmmax=m>n?m:n;
            int mmmin=m<n?m:n;
            int cnt;
            int mmax=-1;
            int i,j,k;
            for(i=1;i<=mmmax;i++)
            {
                cnt=1;
                path[cnt++]=i;
                path[cnt++]=mmmin;
                path[cnt]=mmmax-i;
                memset(hash,0,sizeof(hash));
                memset(vis,0,sizeof(vis));
                if(path[1]>0&&path[2]>0&&path[3]>0)
                {
                    hash[path[1]]=1;hash[path[2]]=1;hash[path[3]]=1;
                    for(j=1;j<=3;j++)
                    {
                        vis[j]=1;
                        dfs(path[j]);
                        vis[j]=0;
                    }
                    int Count=0;
                    for(k=1;k<=mmmin+mmmax;k++)
                    {
                        if(hash[k]==1)
                            Count++;
                    }
                    if(Count>mmax)
                        mmax=Count;
                }
            }
            for(i=1;i<=mmmin;i++)
            {
                cnt=1;
                path[cnt++]=i;
                path[cnt++]=mmmin-i;
                path[cnt]=mmmax;
                memset(hash,0,sizeof(hash));
                memset(vis,0,sizeof(vis));
                if(path[1]>0&&path[2]>0&&path[3]>0)
                {
                    hash[path[1]]=1;hash[path[2]]=1;hash[path[3]]=1;
                    for(j=1;j<=3;j++)
                    {
                        vis[j]=1;
                        dfs(path[j]);
                        vis[j]=0;
                    }
                    int Count=0;
                    for(k=1;k<=mmmin+mmmax;k++)
                    {
                        if(hash[k]==1)
                            Count++;
                    }
                    if(Count>mmax)
                        mmax=Count;
                }
            }
            printf("%d\n",mmax);
        }
        return 0;
    }

  • 相关阅读:
    NHibernate官方文档中文版--拦截器和事件(Interceptors and events)
    NHibernate官方文档中文版——批量插入(Batch inserts)
    NHibernate官方文档中文版-框架架构(Architecture)
    NHibernate官方文档中文版--ISessionFactory的配置(ISessionFactory Configuration)
    ConcurrentDictionary的key判断存在的两种方式,用错了慢100倍
    树莓派3B+,我要跑.NET Core
    C# .NET Framework4.0环境下使用async/await语法,以及其中的需要注意的地方。
    C# 利用结构体对固定格式数据进行解析
    简单工厂模式
    正则表达式替换指定格式内容,实现类似String的Format方法(用扩展方法)
  • 原文地址:https://www.cnblogs.com/zafuacm/p/3089470.html
Copyright © 2011-2022 走看看