[hdu1847]博弈，推理

题意：一堆石子，有n个，两个人轮流取，每次都只能取2的幂次方个数，不能取的人输

思路：首先0是必败态，2的所有幂次都是必胜态。由于选的数模3只能是1或2，恰好又都是2的幂次，0,、3都为必败态，猜想3的所有倍数也为必败态，证明如下：设状态为x=3k，先手任选一个t，那么(x-t)%3不是1就是2，后手就取(x-t)%3，使得先手面临的状态始终是3的倍数，并且只要先手可以取，那么后手也就可以取，所以3的倍数都是必败态。对于x=3k+p，p=1或2的状态，先手都可以将其变成必败态，即先手取p=x%3即可，因此为必胜态。

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/* ******************************************************************************** */ #include <iostream>                                                                 // #include <cstdio>                                                                   // #include <cmath>                                                                    // #include <cstdlib>                                                                  // #include <cstring>                                                                  // #include <vector>                                                                   // #include <ctime>                                                                    // #include <deque>                                                                    // #include <queue>                                                                    // #include <algorithm>                                                                // #include <map>                                                                      // #include <cmath>                                                                    // using namespace std;                                                                //                                                                                     // #define pb push_back                                                                // #define mp make_pair                                                                // #define X first                                                                     // #define Y second                                                                    // #define all(a) (a).begin(), (a).end()                                               // #define fillchar(a, x) memset(a, x, sizeof(a))                                      //                                                                                     // typedef pair<int, int> pii;                                                         // typedef long long ll;                                                               // typedef unsigned long long ull;                                                     //                                                                                     // #ifndef ONLINE_JUDGE                                                                // void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}    // void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>                    // void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;          // while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>      // void print(const T t){cout<<t<<endl;}template<typename F,typename...R>              // void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>   // void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}   // #endif // ONLINE_JUDGE                                                              // template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}        // template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}        // template<typename T>                                                                // void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}            // template<typename T>                                                                // void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}            //                                                                                     // const double PI = acos(-1.0);                                                       // const int INF = 1e9 + 7;                                                            //                                                                                     // /* -------------------------------------------------------------------------------- */   int main() { #ifndef ONLINE_JUDGE     freopen("in.txt", "r", stdin);     //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE     int n;     while (cin >> n) {         puts(n % 3? "Kiki" : "Cici");     }     return 0; } /* ******************************************************************************** */