hdu5795A Simple Nim

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5795

题意：n堆石子，两个人轮流拿，每次可以选择任意一堆取任意个（不能不拿）或者将一个堆分成3个小堆，问先手胜还是后手胜。

分析：打表找sg函数的规律，不会证明。。i=8*k+7时sg[i]=8*k+8，i=8*k+8时sg[i]=8*k+7其他情况sg[i]=i。

代码：

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<vector>#include<string>#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;const int N=1010;const int mod=100000000;const int MOD1=1000000007;const int MOD2=1000000009;const double EPS=0.00000001;typedef long long ll;const ll MOD=1000000007;const int INF=1000000010;const ll MAX=1ll<<55;const double pi=acos(-1.0);typedef double db;typedef unsigned long long ull;int main(){    int a,i,n,t,xo;    scanf("%d", &t);    while (t--) {        scanf("%d", &n);        for (xo=0,i=1;i<=n;i++) {            scanf("%d", &a);            if (a%8==0) xo^=a-1;            else if (a%8==7) xo^=a+1;                else xo^=a;        }        if (xo) printf("First player wins.\n");        else printf("Second player wins.\n");    }    return 0;}