题意：神人鬼三个种族，神只说真话，鬼只说假话，人白天说真话，晚上说假话。根据对话内容区分种族和白天黑夜。  最多有A, B, C, D, E五个人

算法：枚举A, B, C, D, E的种族情况和白天黑夜的情况。每个人有可能是神人鬼，所以用两个bit表示，白天黑夜用1个bit表示。一共需要11 bits.

 

#include
     
      #include
      
       #include
       
        using namespace std;struct statement{    char a;//speaker    char b;//object 等于状态的偏移量    bool c;//not    char d;} p[50];/*A: I am [not] ( divine | human | evil | lying ).A: X is [not] ( divine | human | evil | lying ).A: It is ( day | night ).*/void get(statement &t){    char c;    cin >> c;    t.a = c - 'A';    cin >> c >> c;    if (c == 'I')    {        cin.get(c);        if (c == 't')//It is            t.b = 5;//day|night偏移5        else//I am            t.b = t.a;    }    else        t.b = c - 'A';//0-4对应 A-E  5对应day|night    string s;    cin >> s >> s;    if (s == "not") t.c = true, cin >> s; else t.c = false;    switch (s[0])    {    case 'd': t.d = 0; break;//day divine    case 'e': t.d = 2; break;//evil    case 'l': t.d = 3; break;//lying    default: t.d = 1;//human night    }}//human + night = 2//evil + (day|night) >= 2bool lying(short state, char who){    return (state >> who*2 & 3) + (state >> 10 & 1) >= 2;}//this is too hard to understandbool check(statement st, short s){    return (st.d == 3 ? lying(s, st.b) : st.d == (s >> st.b + st.b & 3)) ^ st.c ^ lying(s, st.a);}//A-E 每个占2位; day,night 占1位const int MAX_STATE=1<<11;bool is_valid_state(int s){    for(int i=0;i<5;i++)        if((s>>i*2 & 3)==3)            return false;    return true;}//this is better then check function//判断说话的人是否说谎话//判断话的内容是否是真的bool judge(statement st, short s){    //if speaker is lying    bool speaker=lying(s, st.a);    //if statement is true    bool object;    //It is ( day | night ).    if(st.b==5)    {        object = (s >> 10 & 1)==st.d;        return speaker!=object;    }    //lying    if(st.d==3)    {        if(st.c)            object = !lying(s, st.b);        else            object = lying(s, st.b);        return speaker!=object;    }    // X is [not] ( divine | human | evil )    object = (s >> st.b*2 & 3)==st.d;    return speaker!=object;}//如果状态矛盾，则置为 11bvoid combine(short &a, short b){    for (int i = 0; i < 6; ++i)        if (a & (3 << i + i) ^ b & (3 << i + i))            a |= (3 << i + i);}const char *s[] = { "divine", "human", "evil" };int main(){#ifndef ONLINE_JUDGE    freopen("./uva592.in", "r", stdin);#endif    int Kase=0;    int n;    //printf("%0X\n", MAX_STATE);    while(cin>>n && n)    {        cout << "Conversation #" << ++Kase << endl;        short a = -1;        for(int i=0; i
        
         >i*2 & 3;            if (x != 3) cout << char(i + 'A') << " is " << s[x] << "." << endl;        }        int x = a>>10 & 3;        if (x != 3) cout << "It is " << (x ? "night." : "day.") << endl;        cout << endl;    }    return 0;}