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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#36825#2447. Domino CoveringNaCly_FishTL 0ms0kbC++6.7kb2022-06-28 22:57:062022-06-28 22:57:07

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-06-28 22:57:07]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:0kb
  • [2022-06-28 22:57:06]
  • 提交

answer

#pragma GCC optimize (2)
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
#include<ctime>
#define ll long long
#define N 10003
#define M 180
#define mid ((l+r)>>1)
using namespace std;

int siz,p;

inline int power(int a,int t){
    int res = 1;
    while(t){
        if(t&1) res = (ll)res*a%p;
        a = (ll)a*a%p;
    	t >>= 1;
    }
    return res;
}

namespace Berlekamp_Massey{
    inline void add(int &x,int y){
        x += y;
        if(x>=p) x -= p;
    }

    inline void dec(int &x,int y){
        x -= y;
        if(x<0) x += p;
    }

    int cnt;
    int Fail[N],delta[N];
    vector<int> R[N];

    int solve(int n,const int *a,int *f){
		int k = 0;
        memset(Fail,0,sizeof(Fail));
        memset(delta,0,sizeof(delta));
        R[0].clear();
        cnt = 0;
        for(int i=1;i<=n;++i){
            if(cnt==0){
                if(a[i]){
                    Fail[cnt++] = i;
                    delta[i] = a[i];
                    R[cnt].resize(0);
                    R[cnt].resize(i,0);
                }
                continue;
            }
            int sum = 0,m = R[cnt].size();
            delta[i] = a[i];
            Fail[cnt] = i;
            for(int j=0;j<m;++j)
                sum = (sum+(ll)a[i-j-1]*R[cnt][j])%p;
            dec(delta[i],sum);    
            if (!delta[i]) continue;
            int id = cnt-1,v = i-Fail[id]+(int)R[id].size();
            for(int j=0;j<cnt;++j){
                if(i-Fail[j]+(int)R[j].size()<v){
                    id = j;
                    v = i-Fail[j]+(int)R[j].size();
                }
            }
            int tmp = (ll)delta[i]*power(delta[Fail[id]],p-2)%p;
            R[cnt+1] = R[cnt];
            while(R[cnt+1].size()<v) R[cnt+1].push_back(0);
            add(R[cnt+1][i-Fail[id]-1],tmp);
            for(int j=0;j<R[id].size();++j)
                dec(R[cnt+1][i-Fail[id]+j],(ll)tmp*R[id][j]%p);
            cnt++;
        }
        k = R[cnt].size();
        for(int i=1;i<=k;++i) f[i] = R[cnt][i-1];
		return k;
    }
}

struct mat{
	int a[36][36];
	inline mat(){ memset(a,0,sizeof(a)); }
};

inline mat operator * (const mat& lhs,const mat& rhs){
	mat res = mat();
	for(int i=0;i<siz;++i)
	for(int j=0;j<siz;++j)
	for(int k=0;k<siz;++k)
		res.a[i][j] = (res.a[i][j]+(ll)lhs.a[i][k]*rhs.a[k][j])%p;
	return res;
}

inline mat operator * (const int& lhs,const mat& rhs){
	mat res = mat();
	for(int i=0;i<siz;++i)
	for(int j=0;j<siz;++j)
		res.a[i][j] = (ll)lhs*rhs.a[i][j]%p;
	return res;
}

inline mat operator + (const mat& lhs,const mat& rhs){
	mat res;
	for(int i=0;i<siz;++i)
	for(int j=0;j<siz;++j)
		res.a[i][j] = (lhs.a[i][j]+rhs.a[i][j])%p;
	return res;
}

inline mat operator - (const mat& lhs,const mat& rhs){
	mat res;
	for(int i=0;i<siz;++i)
	for(int j=0;j<siz;++j)
		res.a[i][j] = (lhs.a[i][j]-rhs.a[i][j]+p)%p;
	return res;
}

inline mat power(mat a,ll t){
	mat res = mat();
	for(int i=0;i<siz;++i) res.a[i][i] = 1;
	while(t){
		if(t&1) res = res*a;
		t >>= 1;
		if(t==0) break;
		a = a*a;
	}
	return res;
}

inline void mod(const int *f,const int *g,int n,int m,int *R){
	static int a[N],b[N],q[N];
	memcpy(a,f,(n+1)<<2),memcpy(b,g,(m+1)<<2);
	for(int i=n-m;~i;--i){
		q[i] = (ll)a[i+m]*power(b[m],p-2)%p;
		for(int j=0;j<=m;++j) 
			a[m+i-j] = (a[m+i-j]-(ll)q[i]*b[m-j])%p;
	}
	for(int i=0;i<m;++i) R[i] = a[i];
}

inline void multiply(const int *f,const int *g,int n,int m,int *r){
	int h[N];
	memset(h,0,(n+m+1)<<2);
	for(int i=0;i<=n;++i)
	for(int j=0;j<=m;++j)
		h[i+j] = (h[i+j]+(ll)f[i]*g[j])%p;
	memcpy(r,h,(n+m+1)<<2);
}

struct complex{ 
    int a,b;
    complex(int a=0,int b=0):a(a),b(b){}

    complex operator + (const complex& x) const{
        return complex((a+x.a)%p,(b+x.b)%p);
    }
    complex operator - (const complex& x) const{
        return complex((a-x.a)%p,(b-x.b)%p);
    }
    complex operator * (const complex& x) const{
        complex res;
        return complex(((ll)a*x.a+5ll*b*x.b)%p,((ll)a*x.b+(ll)b*x.a)%p);
    }
};

inline complex power(complex a,ll t){
    complex res = complex(1,0); 
    while(t){
        if(t&1) res = res*a;
        a = a*a;
        t >>= 1;
    }
    return res;
}

inline int fib(ll n){
	int inv2 = (p+1)>>1;
	complex x = complex(inv2,inv2);
	complex y = complex(inv2,p-inv2);
	complex res = power(x,n)-power(y,n);
	return (res.b+p)%p;
}

int n,ans,lim;
ll m;
int B[36][36],D[36][36],F[M][36][36];
int g[M],cf[M];
int a[M],b[M],res[M];
int dc[36][36];

int main(){
	int T,k;
	scanf("%d",&T);
	while(T--){
		scanf("%d%lld%d",&n,&m,&p);
		if((n&1)&(m&1)){
			puts("0");
			continue;
		}
		if(n==1){
			puts("1");
			continue;
		}
		if(n==2){
			printf("%d\n",fib(m+1));
			continue;
		}
		for(int i=0;i<=n;++i) dc[0][i] = 1;
		for(int i=1;i<=n;++i){
			dc[i][0] = 1;
			for(int j=1;j<=n;++j)
				dc[i][j] = ((dc[i-1][j]+dc[i][j-1])%p+dc[i-1][j-1])%p;
		}
		siz = n,lim = (n<<1)+2;		
		memset(F[0],0,sizeof(F[0]));
		memset(F[1],0,sizeof(F[1]));
		for(int i=0;i<n;++i) F[0][i][i] = 1;
		for(int i=0;i<(n-1);++i){
			F[1][i][i+1] = 1;
			F[1][i+1][i] = p-1;
		}
		for(int i=2;i<=lim;++i){
			for(int j=0;j<n;++j)
			for(int k=0;k<n;++k){
				F[i][j][k] = p-F[i-2][j][k];
				if(k-1>=0) F[i][j][k] = (F[i][j][k]+F[i-1][j][k-1])%p;
				if(k+1<n) F[i][j][k] = (F[i][j][k]-F[i-1][j][k+1]+p)%p;
			}
		}
		k = (n<<1);
		for(int i=1;i<=k;++i){
			int ti = i>>1;
			cf[i] = (i&1)?-dc[ti][n-ti]:0;
		}
		reverse(cf+1,cf+k+1);
		memset(a,0,(k+1)<<3),memset(b,0,(k+1)<<3);
		memset(res,0,(k+1)<<3);
		for(int i=1;i<=k;++i) a[k-i] = p-cf[i];
		a[k] = b[1] = res[0] = 1;
		int bt = 1,rt = 0;
		ll tm = m;
		while(1){
			if(m&1){
				multiply(res,b,rt,bt,res);
				rt += bt;
				if(rt>=k) mod(res,a,rt,k,res);
				rt = min(rt,k-1);
			}
			m >>= 1;
			if(m==0) break;
			multiply(b,b,bt,bt,b);
			bt <<= 1;
			if(bt>=k) mod(b,a,bt<<1,k,b);
			bt = min(bt,k-1);
		}
		memset(D,0,sizeof(D));
		for(int i=0;i<k;++i)
		for(int t=0;t<n;++t)
		for(int j=0;j<n;++j)
			D[t][j] = (D[t][j]+(ll)res[i]*F[i][t][j]%p+p)%p;
		for(int i=0;i<n;++i){
			if((i&1)!=(n&1)) continue;
			for(int j=0;j<n;++j){
				if(D[i][j]==0) continue;
				B[i>>1][j>>1] = D[i][j];
			}
		}
		n >>= 1;
		int d;
		for(int i=0;i<n;++i)
		for(int j=i+1;j<n;++j){
			if(B[i][j]==0) continue;
			d = (ll)B[j][i]*power(B[i][i],p-2)%p;
			for(int k=i;k<n;++k) B[j][k] = (B[j][k]-(ll)d*B[i][k])%p;
		}
		ans = 1;
		for(int i=0;i<n;++i) ans = (ll)ans*B[i][i]%p;
		//if(((__int128_t)n*tm/2)%2==1) ans = p-ans;
		ans = (ans%p+p)%p;
		printf("%d\n",ans);
	}
	return 0;
}

详细

Test #1:

score: 0
Time Limit Exceeded

input:

20000
3 695860418 1064851577
5 909984642 1024590071
2 702478034 1015656679
3 832070346 1020170803
3 931276816 1069777147
5 624464668 1019025517
4 563777828 1039054439
3 70355912 1062629389
2 538334151 1043751551
4 644616259 1051984399
2 565963832 1050482821
3 489913670 1030290631
5 625001688 6518147...

output:

889215569
670951730
483405543
355257897
563953115
524024279
815773299
734679810
266201944
156563661
556277524
188969274
607188444
210143750
108744246
7651374
636855097
493227881
66494941
541528569
232780855
220305284
157636415
735388230
99671148
863972210
540368496
249186290
797066419
1022449349
786...

result: