【HDOJ1847】Good Luck in CET-4 Everybody!(博弈)
2016-08-06 16:02
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记录一个菜逼的成长。。
简单博弈 暴力打表
只能取2的幂次
简单博弈 暴力打表
只能取2的幂次
#include <cstdio> #include <iostream> #include <cstring> #include <string> #include <algorithm> #include <cstdlib> #include <vector> #include <set> #include <map> #include <queue> #include <list> #include <deque> #include <cctype> #include <bitset> using namespace std; #define cl(a) memset(a,0,sizeof(a)) typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> PII; const int INF = 0x3f3f3f3f; const int maxn = 1000 + 10; int sg[maxn],two[12]; bool Hash[maxn]; void init() { two[0] = 1; for( int i = 1; i < 10; i++ ){ two[i] = two[i-1] << 1; } } void sg_solve(int n,int m) { init(); memset(sg,0,sizeof(sg)); for( int j,i = 1; i <= n; i++ ){ memset(Hash,0,sizeof(Hash)); for( j = 0; j <= m; j++ ){ if(i - two[j] >= 0)Hash[sg[i-two[j]]] = 1; } for( j = 0; j <= n; j++ ) if(!Hash[j])break; sg[i] = j; } } int main() { sg_solve(maxn-1,9); int n; while(~scanf("%d",&n)){ printf("%s\n",sg ?"Kiki":"Cici"); } return 0; }
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