bzoj3576[Hnoi2014]江南乐

Posted 2020-06-19 maijing

http://www.lydsy.com/JudgeOnline/problem.php?id=3576

SG函数

我们发现$\left \lfloor \frac{n}{i} \right \rfloor$只有$\sqrt{n}$个取值

不妨设$x=\left \lfloor \frac{n}{i} \right \rfloor$

那么最终的分割方案一定是下面4种情况之一：偶数个$x$，偶数个$x+1$；偶数个$x$，奇数个$x+1$；奇数个$x$，偶数个$x+1$；奇数个$x$，奇数个$x+1$。

分类讨论解不定方程即可。

如果判断偶数个$x$，奇数个$x+1$的方案是否存在就是问以下的不定方程有没有解：

$2k_1x+(2k_2+1)(x+1)=n$

$2k_1+2k_2+1\geq 2$

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<utility>
#include<set>
#include<bitset>
#include<vector>
#include<functional>
#include<deque>
#include<cctype>
#include<climits>
#include<complex>
#include<cassert> 
//#include<bits/stdc++.h>适用于CF,UOJ，但不适用于poj
 
using namespace std;

typedef long long LL;
typedef double DB;
typedef pair<int,int> PII;
typedef pair<DB,DB> PDD;
typedef complex<DB> CP;
typedef vector<int> VI;

#define mmst(a,v) memset(a,v,sizeof(a))
#define mmcy(a,b) memcpy(a,b,sizeof(a))
#define fill(a,l,r,v) fill(a+l,a+r+1,v)
#define re(i,a,b)  for(i=(a);i<=(b);i++)
#define red(i,a,b) for(i=(a);i>=(b);i--)
#define fi first
#define se second
#define mp(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define SF scanf
#define PF printf
#define two(k) (1<<(k))
#define SZ(x) (int(x.size()))
#define all(x) (x).begin(),(x).end()
#define ire(i,v,x) for(i=0,v=i<SZ(x)?x[i]:0;i<SZ(x);v=x[++i])


template<class T>inline T sqr(T x){return x*x;}
template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;}
template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}

inline int sgn(DB x){if(abs(x)<1e-9)return 0;return(x>0)?1:-1;}
const DB Pi=acos(-1.0);

int gint()
  {
        int res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!=‘-‘ && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z==‘-‘){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-‘0‘,z=getchar());
        return (neg)?-res:res; 
    }
LL gll()
  {
      LL res=0;bool neg=0;char z;
        for(z=getchar();z!=EOF && z!=‘-‘ && !isdigit(z);z=getchar());
        if(z==EOF)return 0;
        if(z==‘-‘){neg=1;z=getchar();}
        for(;z!=EOF && isdigit(z);res=res*10+z-‘0‘,z=getchar());
        return (neg)?-res:res; 
    }

const int maxn=100100;
const int maxm=110;

int F;
int n,SG[maxn];
int m,stone[maxm];
int ci,bak[maxn];

int check(int x,int n,int f)
  {
      if(n<0)return 0;
      if(n&1)return 0;
      n>>=1;
      if(!f) return (n+x+1-1)/(x+1)<=n/x;
      if(n%(x+1)==0 && n/(x+1)>=1)return 1;
      if(n%x==0 && n/x>=1)return 1;
      return (1+n+x+1-1)/(x+1)<=(n-1)/x;
  }

int main()
  {
      freopen("game.in","r",stdin);
      freopen("game.out","w",stdout);
      int i,T=gint();F=gint();
      while(n+1<F)SG[++n]=0;
      while(T--)
        {
            int n2=0;
            m=gint();
            re(i,1,m)upmax(n2,stone[i]=gint());
            while(n<n2)
              {
                  n++;ci++;
                  int x;
                  for(i=2;i<=n;i=n/x+1)
                    {
                        x=n/i;
                        if(check(x,n,1))
                                  bak[0]=ci;
                        if(check(x,n-x,1))
                                  bak[SG[x]]=ci;
                        if(check(x,n-x-1,1))
                                  bak[SG[x+1]]=ci;
                        if(check(x,n-x-x-1,0))
                                  bak[SG[x]^SG[x+1]]=ci;
                    }
                  for(i=0;bak[i]==ci;i++);
                  SG[n]=i;
              }
            int ans=0;
            re(i,1,m)ans^=SG[stone[i]];
            PF("%d ",min(ans,1));
        }
      return 0;
  }

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