Codeforces 305E Playing with String

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传送门：http://codeforces.com/problemset/problem/305/E

思路：首先每个字符都可选的子串可以看成一个独立的游戏，那么最终的答案就可以通过所有独立游戏的SG值的异或和得到。

代码：

#include<cstdio>#include<cstring>#include<algorithm>const int n=5010;using namespace std;int sg[n];char s[n];bool bo[n];int getsg(int len){if (sg[len]!=-1) return sg[len];memset(bo,0,sizeof(bo));bo[getsg(len-2)]=1;for (int i=1;i+i<len;i++) bo[getsg(i-1)^getsg(len-i-2)]=1;for (int i=0;i<n;i++) if (!bo[i]) return sg[len]=i;}int getans(int l,int r){int sum=0;for (int i=l+1;i<=r-1;i++) if (s[i+1]==s[i-1]){int len=0;while (s[i+1]==s[i-1]&&i<=r-1) i++,len++;sum^=sg[len];printf("%d\n",len);}return sum;}int main(){memset(sg,-1,sizeof(sg));sg[0]=0,sg[1]=1;for (int i=2;i<n;i++) if (sg[i]==-1) sg[i]=getsg(i);for (int i=0;i<=10;i++) { printf("%d",sg[i]); }while (scanf("%s",s)!=EOF){bool sec=1;for (int i=1,len=strlen(s);i<len-1;i++){if (s[i-1]!=s[i+1]) continue;if (getans(0,i-1)^getans(i+1,len-1)) continue;printf("First\n%d\n",i+1),sec=0;break;}if (sec) puts("Second");}return 0;}

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