• BZOJ 1228 E&G(sg函数+找规律)


    把一对石子堆看出一个子游戏。打出子游戏的sg表找规律。。

    这个规律我是一定找不出来的。。。

    对于i,j,如果
    (i-1)%pow(2,k+1) < pow(2,k)
    (j-1)%pow(2,k+1) < pow(2,k)
    那么最小的k值就是sg值。
     
    # include <cstdio>
    # include <cstring>
    # include <cstdlib>
    # include <iostream>
    # include <vector>
    # include <queue>
    # include <stack>
    # include <map>
    # include <set>
    # include <cmath>
    # include <algorithm>
    using namespace std;
    # define lowbit(x) ((x)&(-x))
    # define pi acos(-1.0)
    # define eps 1e-9
    # define MOD 1000000000
    # define INF 1000000000
    # define mem(a,b) memset(a,b,sizeof(a))
    # define FOR(i,a,n) for(int i=a; i<=n; ++i)
    # define FO(i,a,n) for(int i=a; i<n; ++i)
    # define bug puts("H");
    # define lch p<<1,l,mid
    # define rch p<<1|1,mid+1,r
    # define mp make_pair
    # define pb push_back
    typedef pair<int,int> PII;
    typedef vector<int> VI;
    # pragma comment(linker, "/STACK:1024000000,1024000000")
    typedef long long LL;
    int Scan() {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    void Out(int a) {
        if(a<0) {putchar('-'); a=-a;}
        if(a>=10) Out(a/10);
        putchar(a%10+'0');
    }
    const int N=20005;
    //Code begin...
    
    int a[N];
    
    int sg(int x, int y){
        LL tmp=2;
        for (int i=0; ; ++i, tmp<<=1) if ((x-1)%tmp<tmp/2 && (y-1)%tmp<tmp/2) return i;
    }
    int main ()
    {
        int T, n;
        scanf("%d",&T);
        while (T--) {
            int ans=0;
            scanf("%d",&n);
            FOR(i,1,n) scanf("%d",a+i);
            FOR(i,1,n/2) ans^=sg(a[2*i-1],a[2*i]);
            puts(ans?"YES":"NO");
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/lishiyao/p/6763706.html
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