Codeforces Gym 100610 Problem H. Horrible Truth 瞎搞

Problem H. Horrible Truth

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100610

Description

In a Famous TV Show “Find Out” there are n characters and only one Horrible Truth. To make the series breathtaking all way long, the screenplay writer decided that every episode should show exactly one important event. There are three types of the important events in this series: • character A finds out the Truth; • character A finds out that the other character B knows the Truth; • character A finds out that the other character B doesn’t know the Truth. Initially, nobody knows the Truth. All events must be correct, and every fact found out must be true. If some character finds out some fact, she cannot find it out once again. Moreover, to give the audience some sense of action, the writer does not want an episode to show the important event of the same type as in the previous episode. Your task is to determine the maximal possible number of episodes in the series and to create an example of a screenplay plan.

Input

The only line of the input contains a single integer n — the number of characters in the TV show (1 ≤ n ≤ 100).

Output

In the first line of the output file output a single integer k — the maximal possible number of episodes in the series. Then write k lines, each containing a description of an episode. For the episode in which character A (characters are numbered 1 through n) finds out the Truth, write the line “A 0”. For an episode in which character A finds out that character B knows the Truth, write the line “A B”. Similarly, for an episode in which character A finds out that character B doesn’t know the Truth, write the line “A -B”. If there are several plans providing the maximal possible number of episodes, output any one of them.

Sample Input

Sample Output

13 2 -1 1 0 2 1 1 -2 3 1 3 -2 2 0 1 2 1 -3 3 2 2 -3 3 0 1 3

HINT

题意

无良电视剧公司要来拍电视剧了

每集演一个故事，分别是叫做A知道，B知道A知道,B知道A不知道

要求每个人都不能在两集内做同样的事情

不能连续两集都是知道，或者知道别人知道，或者知道别人不知道

要求你合乎逻辑的情况下，最多演多少集

题解：

观察样例可以，第一回合嘲讽一下别人，第二回合就让人知道，然后就来回知道别人知道和知道别人不知道这样就好了……

瞎搞……

代码：

//qscqesze

#include <cstdio>

#include <cmath>

#include <cstring>

#include <ctime>

#include <iostream>

#include <algorithm>

#include <set>

#include <bitset>

#include <vector>

#include <sstream>

#include <queue>

#include <typeinfo>

#include <fstream>

#include <map>

#include <stack>

typedef long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define maxn 110000

#define mod 1001

#define eps 1e-9

#define pi 3.1415926

int Num;

//const int inf=0x7fffffff;   //§ß§é§à§é¨f§³

const ll inf=;

inline ll read()

{

    ll x=,f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

//*************************************************************************************

map< pair<int,int> ,int> H;

vector<pair<int,int> > V;

int main()

{ // horrible.out

    freopen("horrible.in","r",stdin);

    freopen("horrible.out","w",stdout);

    int n=read();

    int tot=;

    if(n==)

    {

        printf("1\n");

        printf("1 0\n");

        return ;

    }

    V.push_back(make_pair(,-));

    for(int i=;i<=n;i++)

    {

        V.push_back(make_pair(i,));

        int ttt = ;

        int flag  = ;

        while(flag)

        {

            flag=;

            ttt++;

            if(ttt%==)

            {

                flag=;

                for(int ii=;ii<=n;ii++)

                {

                    if(i==ii)

                        continue;

                    if(H[make_pair(ii,i)])

                        continue;

                    H[make_pair(ii,i)]=;

                    V.push_back(make_pair(ii,i));

                    flag=;

                    break;

                }

            }

            else

            {

                flag=;

                for(int jj=i+;jj<=n;jj++)

                {

                    for(int ii=;ii<=n;ii++)

                    {

                        if(ii==jj)

                            continue;

                        if(H[make_pair(ii,-jj)])

                            continue;

                        H[make_pair(ii,-jj)]=;

                        V.push_back(make_pair(ii,-jj));

                        flag=;

                        break;

                    }

                    if(flag)

                        break;

                }

            }

        }

    }

    printf("%d\n",V.size());

    for(int i=;i<V.size();i++)

        printf("%d %d\n",V[i].first,V[i].second);

}