sdut 3253 Game! 博弈

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题意:

一个1~n的环每个人每次可以拿一个石头或者拿连续2个的石头(即第i个和第i+1个)

最后一个取完的就是赢家问先手是否能胜

分析:

当n为1和2的时候先手必胜,当n为3的时候为必败当n为4的时候无论怎么取都将进入必胜态即n为4先手必败,手推几个情况后断定n大于2的时候先手必败

ACcode:

#pragma warning(disable:4786)//使命名长度不受限制#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈#include <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rd3(x,y,z) scanf("%d%d%d,&x,&y,&z)#define rdl(x) scanf("%I64d,&x);#define rds(x) scanf("%s",x)#define rdc(x) scanf("%c",&x)#define ll long long int#define ull unsigned long long#define maxn 1005#define mod 1000000007#define INF 0x3f3f3f3f //int 最大值#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI  acos(-1.0)#define E  exp(1)#define eps 1e-8ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}ll mul(ll a,ll b,ll p){ll sum=0;for(;b;a=(a+a)%p,b>>=1)if(b&1)sum=(sum+a)%p;return sum;}inline void Scan(int &x) {      char c;while((c=getchar())<'0' || c>'9');x=c-'0';      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';}using namespace std;int main(){    ull n;    int loop;    scanf("%d",&loop);    while(loop--){        cin>>n;        if(n>=3)puts("blankcqk");        else puts("zbybr");    }    return 0;}


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