• Arithmetic Progressions chapter 1.4


      感觉蛮简单,就是枚举,以为能一次过,结果到case7直接tle了,自己修改了下,也去网上看了下解题报告,尝试剪枝,最后到把set<int> si,si.count()换成用一个数组判断存在才过,看来set虽然方便,可还得少用...

    /*
    
    ID: hubiao cave
    
    PROG: ariprog
    
    LANG: C++
    
    */
    
    
    
    
    #include<iostream>
    #include<fstream>
    #include<string>
    #include<set>
    
    using namespace std;
    
    set<int> si;
    //set<int> ssi;
    
    struct info
    {
        int a;
        int b;
    };
    
    
    
    
    set<info>sin;
    
    
    
    bool operator < (const info&,const info&);
    bool operator ==(const info&,const info&);
    int main()
    
    {
    
        ifstream fin("ariprog.in");
        ofstream fout("ariprog.out");
        int upper;
        int number;
        int limit;
    
        fin>>number>>upper;
        limit=2*upper*upper;
    
        int* ary=new int [limit+1];
    
        for(int i=0;i<=limit;i++)
        {
            ary[i]=0;
        }
    
        int ul=limit/(number-1);
        
        for(int i=0;i<=upper;i++)
            for(int j=0;j<=upper;j++)
        {
            si.insert(i*i+j*j);
            ary[i*i+j*j]=1;
        }
    
            for(set<int>::iterator it=si.begin();it!=si.end();it++)
            {
                int now=*it;
                int tnow=now;
                int count=1;
    
                for(int i=1;i<=ul;i++)
                {
                    
                    info te;
                    te.a=now;
                    te.b=i;
    
                    if(sin.count(te))
                        continue;
                    if(now+i*(number-1)>limit)
                        break;
                    //while(si.count(now+i))
                    while((now+i<=limit)&&ary[now+i])
                    {
                        
                        now=now+i;
                        count++;
                    }
    
                    if(count==number)
                    {
                        info in;
                        in.a=tnow;
                        in.b=i;
                        sin.insert(in);
                        //rec.insert(in);
                    }
    
                    if(count>number)
                    {
                        int n=count-number;
                        for(int m=0;m<=n;m++)
                        {
                            info in;
                            in.a=tnow+i*m;
                            in.b=i;
                            sin.insert(in);
                            //rec.insert(in);
                        }
    
                  }
                    count=1;
                    now=*it;
    
                    tnow=now;
    
                }
            }
    
            if(sin.size()==0)
            {
                fout<<"NONE"<<endl;
                return 0;
            }
            for(set<info>::iterator it=sin.begin();it!=sin.end();it++)
            {
                fout<<it->a<<" "<<it->b<<endl;
            }
        
    
        return 0;
    
    
    }
    
    
    bool operator < (const info& i1,const info& i2)
    {
        if(i1.b!=i2.b)
            return i1.b<i2.b;
        else
            return i1.a<i2.a;
    }
    
    bool operator ==(const info& i1,const info& i2)
    {
        if(i1.a==i2.a&&i1.b==i2.b)
            return true;
        return false;
    }
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  • 原文地址:https://www.cnblogs.com/cavehubiao/p/3257491.html
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