HDU 1847 Good Luck in CET-4 Everybody!
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传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1847
题目大意:中文题目不解释
分析:
根据必胜点和必败点的定义就可以很好地做掉这个题了。做个必胜点和必败点的标记就行了。没有标记过的点为必败点,从必败点一步到达的点都标记为必胜点。
代码
#include <iostream>using namespace std;int a[10]={1,2,4,8,16,32,64,128,256,512};int main(){ int f[1005]={1,1,1},i,j,n; for(i=3;i<1001;i++) { if(!f[i]) for(j=0;j<10 && i+a[j]<1001;j++) f[i+a[j]] = 1; } while(cin >> n) if(f[n]) cout << "Kiki" << endl; else cout << "Cici" << endl; return 0;}
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