Codeforces Round #334 (Div. 1) C. Lieges of Legendre
题意:有n堆牛,每堆有ai头牛。两个人玩一个游戏,游戏规则为:
<1>从任意一个非空的堆中移走一头牛;
<2>将偶数堆2*x变成k堆,每堆x头牛(可以增加牛的个数)
移走最后一头牛的人获胜;
数据:n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109).a1, a2, ... an (1 ≤ ai ≤ 109)
分析:这显然是SG函数与SG定理的应用;
SG函数与SG定理:SG[x] = mex(S);其中S是所有x的后继状态的SG函数值的集合,mex(S) 表示不在S内的最小非负整数;SG[x] = 0当且仅当x为必败状态;
游戏和的SG函数等于各子游戏的SG函数的nim和(nim和就是异或操作)
理解:对于初始的状态要求出SG函数值,就必须等到所有的子状态的SG函数,但是由于ai的值太大(是对操作1而言的),这是一般需要打表找出规律;即数组ret的两维表示k偶奇时前5个值的SG值,并且容易发现不论k的奇偶,当i >= 5 且 i为奇数时,ai = 0;
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<cstdio> #include<cstring> #include<string.h> #include<algorithm> #include<vector> #include<cmath> #include<stdlib.h> #include<time.h> #include<stack> #include<set> #include<map> #include<queue> using namespace std; #define rep0(i,l,r) for(int i = (l);i < (r);i++) #define rep1(i,l,r) for(int i = (l);i <= (r);i++) #define rep_0(i,r,l) for(int i = (r);i > (l);i--) #define rep_1(i,r,l) for(int i = (r);i >= (l);i--) #define MS0(a) memset(a,0,sizeof(a)) #define MS1(a) memset(a,-1,sizeof(a)) #define MSi(a) memset(a,0x3f,sizeof(a)) #define inf 0x3f3f3f3f #define lson l, m, rt << 1 #define rson m+1, r, rt << 1|1 typedef __int64 ll; template<typename T> void read1(T &m) { T x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} m = x*f; } template<typename T> void read2(T &a,T &b){read1(a);read1(b);} template<typename T> void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);} template<typename T> void out(T a) { if(a>9) out(a/10); putchar(a%10+'0'); } int k; int ret[][5] = {{0,1,2,0,1},{0,1,0,1,2}}; int mex(int state) { if(state < 5) return ret[k%2][state];// G[0] = 0,G[1] = 1; int a,b; if(state & 1){ return 0; }else{ if(k & 1) b = mex(state/2); else b = 0; a = 0;// mex(state - 1) = 0; if(a != 0 && b != 0) return 0; if(a != 1 && b != 1) return 1; return 2; } } int main() { int n,ans = 0,x; read2(n,k); rep1(i,1,n){ read1(x); //ans = 0; ans ^= mex(x);//mex(i)用来找规律; //cout<<ans<<" "; } printf("%s",ans?"Kevin":"Nicky"); return 0; }