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。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 | /* ******************************************************************************** */#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;}/* ******************************************************************************** */ |