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## Re:最简单的想法

Posted by 20101003712 at 2011-11-09 15:22:16 on Problem 1176
In Reply To:最简单的想法 Posted by:hehexiaobai at 2010-08-22 20:53:30
```> 每种翻法有0或1，分别代码不翻转和翻转。（翻两次又回到原来的状态）
>
> 因此可以枚举，一共16中状态。附下丑陋的代码
>
> #include<iostream>
> #include<string>
> #include<algorithm>
> #include<vector>
> using namespace std;
> int n,c,t;
> bool f[101];
> bool g[101];
> int last[101];
>
> bool judge()
> {
>      for(int i=1; i<=n; i++)
>              if(last[i]!=-1&&last[i]!=g[i])
>               return false;
>      return true;
> }
>
> void turn1()
> {
>      for(int i=1; i<=n; i++)
>            g[i]=!g[i];
> }
>
> void turn2()
> {
>      for(int i=1; i<=n; i+=2)
>           g[i]=!g[i];
> }
>
> void turn3()
> {
>      for(int i=2; i<=n;i+=2)
>              g[i]=!g[i];
> }
>
> void turn4()
> {
>
>      for(int i=0,k=1; k<=n;)
>      {
>            g[k]=!g[k];
>            i++;
>            k=3*i+1;
>      }
> }
>
> int main()
> {
>     cin>>n;
>     cin>>c;
>     vector<string> vec;
>     string s;
>     for(int i=1; i<=n; i++)f[i]=1;
>     for(int i=1; i<=n; i++)last[i]=-1;
>     while(cin>>t&&t!=-1)
>             last[t]=1;
>     while(cin>>t&&t!=-1)
>           last[t]=0;
>
>     int i1, i2, i3, i4,i;
>     for(i1=0; i1<=1; i1++)
>     for(i2=0; i2<=1; i2++)
>     for(i3=0; i3<=1; i3++)
>     for(i4=0; i4<=1; i4++){
>               if( c>=(i1+i2+i3+i4)&&(c-(i1+i2+i3+i4))%2==0 )
>               {
>                       for(i=1; i<=n;i++)
>                               g[i]=f[i];
>                       if(i1)turn1();
>                       if(i2)turn2();
>                       if(i3)turn3();
>                       if(i4)turn4();
>
>                       if(judge()){
>                            for(s="",i=1; i<=n; i++)
>                                  s+='0'+g[i];
>                            vec.push_back(s);
>                       }
>               }
>     }
>
>     sort(vec.begin(),vec.end());
>     if(vec.size()==0)cout<<"IMPOSSIBLE"<<endl;
>     else{
>        for(i=0; i<vec.size(); i++)
>                 cout<<vec[i]<<endl;
>        }
>
>     return 0;
> }

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