LOJ6436. 「PKUSC2018」神仙的游戏 [NTT]
作者:互联网
思路
首先通过各种手玩/找规律/严谨证明,发现当\(n-i\)为border当且仅当对于任意\(k\in[0,i)\),模\(i\)余\(k\)的位置没有同时出现0和1。
换句话说,拿出任意一个1的位置\(x\),一个0的位置\(y\),那么对于\(|x-y|\)的所有约数\(i\),\(n-i\)均不合法。
考虑用NTT优化这个过程:记两个多项式\(A(x),B(x)\)。若\(s_i=0\)则\([x^i]A(x)=1\);若\(s_i=1\)则\([x^{n-i}]B(x)=1\)。然后把\(A\)和\(B\)卷积起来即可。
代码
#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
using namespace std;
#define pii pair<int,int>
#define fir first
#define sec second
#define MP make_pair
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define drep(i,x,y) for (int i=(x);i>=(y);i--)
#define go(x) for (int i=head[x];i;i=edge[i].nxt)
#define templ template<typename T>
#define sz 4004040
#define mod 998244353ll
typedef long long ll;
typedef double db;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
templ inline void read(T& t)
{
t=0;char f=0,ch=getchar();double d=0.1;
while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
t=(f?-t:t);
}
template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
char __sr[1<<21],__z[20];int __C=-1,__zz=0;
inline void Ot(){fwrite(__sr,1,__C+1,stdout),__C=-1;}
inline void print(register int x)
{
if(__C>1<<20)Ot();if(x<0)__sr[++__C]='-',x=-x;
while(__z[++__zz]=x%10+48,x/=10);
while(__sr[++__C]=__z[__zz],--__zz);__sr[++__C]='\n';
}
void file()
{
#ifdef NTFOrz
freopen("a.in","r",stdin);
#endif
}
inline void chktime()
{
#ifndef ONLINE_JUDGE
cout<<(clock()-t)/1000.0<<'\n';
#endif
}
#ifdef mod
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
ll inv(ll x){return ksm(x,mod-2);}
#else
ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
#endif
// inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;
int r[sz],limit;
void NTT_init(int n)
{
limit=1;int l=-1;
while (limit<=n+n) limit<<=1,++l;
rep(i,0,limit-1) r[i]=(r[i>>1]>>1)|((i&1)<<l);
}
void NTT(ll *a,int type)
{
rep(i,0,limit-1) if (i<r[i]) swap(a[i],a[r[i]]);
for (int mid=1;mid<limit;mid<<=1)
{
ll Wn=ksm(3,(mod-1)/mid>>1);if (type==-1) Wn=inv(Wn);
for (int len=mid<<1,j=0;j<limit;j+=len)
{
ll w=1;
for (int k=0;k<mid;k++,w=w*Wn%mod)
{
ll x=a[j+k],y=a[j+k+mid]*w%mod;
a[j+k]=(x+y)%mod;a[j+k+mid]=(x-y+mod)%mod;
}
}
}
if (type==1) return;
ll I=inv(limit);
rep(i,0,limit-1) a[i]=a[i]*I%mod;
}
int n;
char s[sz];
ll tmp1[sz],tmp2[sz],a[sz];
ll ans;
int main()
{
file();
cin>>(s+1);n=strlen(s+1);
rep(i,1,n) if (s[i]=='0') tmp1[i]=1;
rep(i,1,n) if (s[i]=='1') tmp2[n-i]=1;
NTT_init(n);
NTT(tmp1,1);NTT(tmp2,1);
rep(i,0,limit-1) tmp1[i]=tmp1[i]*tmp2[i]%mod;
NTT(tmp1,-1);
rep(i,1,n+n) a[i]=tmp1[i];
rep(i,1,n-1)
{
bool flg=1;
for (int j=i;j<n;j+=i) flg&=(a[n-j]==0&&a[n+j]==0);
if (flg) ans^=1ll*(n-i)*(n-i);
}
ans^=1ll*n*n;
cout<<ans;
return 0;
}
标签:ch,PKUSC2018,NTT,ret,LOJ6436,mod,tmp1,ll,define 来源: https://www.cnblogs.com/p-b-p-b/p/10869241.html