HDU 1847 Good Luck in CET-4 Everybody!
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题目:
http://acm.hdu.edu.cn/showproblem.php?pid=1847
题解:
同样的思维,我竟没想出来。http://blog.csdn.net/yl_freedom/article/details/20796733
以后做题都要有从简单的情况入手的思维,就这题来说,我们立马可以想到的一种必输情况是3。往前一步,当n为3的倍数时,无论先取者取多少,另一个人都可以取1张或2张使剩余的牌还是3的倍数。所以当n%3==0时,先取牌者必输。
代码:
#include<stdio.h>int main(){int n;while(~scanf("%d",&n)){if(n%3==0) printf("Cici\n");else printf("Kiki\n");}return 0;}
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