QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#369249#6323. Range NEQVengeful_Spirit#RE 24ms14872kbC++175.0kb2024-03-27 22:31:312024-03-27 22:31:32

Judging History

你现在查看的是最新测评结果

  • [2024-03-27 22:31:32]
  • 评测
  • 测评结果:RE
  • 用时:24ms
  • 内存:14872kb
  • [2024-03-27 22:31:31]
  • 提交

answer

#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
using namespace std;
// void cmax(int &x, int y) {x  = std::max(x, y);}
typedef long long ll;
const int mod=998244353;
inline int inc(int x,int y){return x+y>=mod?x+y-mod:x+y;}
inline int dec(int x,int y){return x-y<0?x-y+mod:x-y;}
inline int mul(int x,int y){return (ll)x*y%mod;}
inline int qpow(int x,int y){
	int res=1;
	for(;y;y>>=1)res=y&1?mul(res,x):res,x=mul(x,x);
	return res;
}
inline int inv(int x){return qpow(x,mod-2);}

const int N=1<<20;
namespace NTT{
int re[N],w[2][N];
inline int getre(int n){
	int len=1,bit=0;
	while(len<n)len<<=1,++bit;
	for(int i=1;i<len;++i)re[i]=(re[i>>1]>>1)|((i&1)<<(bit-1));
	w[0][0]=w[1][0]=1,w[0][1]=qpow(3,(mod-1)/len),w[1][1]=inv(w[0][1]);
	for(int o=0;o<2;++o)for(int i=2;i<len;++i)
		w[o][i]=mul(w[o][i-1],w[o][1]);
	return len;
}
inline void NTT(int* a,int n,int o=0){
	for(int i=1;i<n;++i)if(i<re[i])swap(a[i],a[re[i]]);
	int R;
	for(int k=1;k<n;k<<=1)
		for(int i=0,t=k<<1,st=n/t;i<n;i+=t)
			for(int j=0,nw=0;j<k;++j,nw+=st)
				R=mul(a[i+j+k],w[o][nw]),a[i+j+k]=dec(a[i+j],R),a[i+j]=inc(a[i+j],R);
	if(o){
		R=inv(n);
		for(int i=0;i<n;++i)a[i]=mul(a[i],R);
	}
}

int t0[N],t1[N],t2[N];
inline void multi(const int* a,const int* b,int* c,int n,int m){
	if(n+m<=32){
		memset(c,0,sizeof(int)*(n+m+1));
		for(int i=0;i<=n;++i)
			for(int j=0;j<=m;++j)
				c[i+j]=inc(c[i+j],mul(a[i],b[j]));
		return;
	}
	int len=getre(n+m+1);
	memset(t0,0,sizeof(int)*len),memcpy(t0,a,sizeof(int)*(n+1));
	memset(t1,0,sizeof(int)*len),memcpy(t1,b,sizeof(int)*(m+1));
	NTT(t0,len),NTT(t1,len);
	for(int i=0;i<len;++i)t0[i]=mul(t0[i],t1[i]);
	NTT(t0,len,1);
	memcpy(c,t0,sizeof(int)*(n+m+1));
}
inline void inver(const int* a,int* b,int n){
	int len=1;
	while(len<=n)len<<=1;
	memset(t0,0,sizeof(int)*len),memcpy(t0,a,sizeof(int)*(n+1));
	memset(t1,0,sizeof(int)*(len<<1));
	memset(t2,0,sizeof(int)*(len<<1));
	t2[0]=inv(t0[0]);
	for(int k=1;k<=len;k<<=1){
		memcpy(t1,t0,sizeof(int)*k);
		getre(k<<1);
		NTT(t1,k<<1),NTT(t2,k<<1);
		for(int i=0;i<(k<<1);++i)t2[i]=mul(t2[i],dec(2,mul(t1[i],t2[i])));
		NTT(t2,k<<1,1);
		for(int i=k;i<(k<<1);++i)t2[i]=0;
	}
	memcpy(b,t2,sizeof(int)*(n+1));
}
} //namespace NTT

struct poly:public vector<int>{
	inline int time()const{return size()-1;}
	inline poly(int tim=0,int c=0){
		resize(tim+1);
		if(tim>=0)at(0)=c;
	}
	inline poly operator%(const int& n)const{
		poly r(*this);
		r.resize(n);
		return r;
	}
	inline poly operator%=(const int& n){
		resize(n);
		return *this;
	}
	inline poly operator+(const poly& p)const{
		int n=time(),m=p.time();
		poly r(*this);
		if(n<m)r.resize(m+1);
		for(int i=0;i<=m;++i)r[i]=inc(r[i],p[i]);
		return r;
	}
	inline poly operator-(const poly& p)const{
		int n=time(),m=p.time();
		poly r(*this);
		if(n<m)r.resize(m+1);
		for(int i=0;i<=m;++i)r[i]=dec(r[i],p[i]);
		return r;
	}
	inline poly operator*(const poly& p)const{
		poly r(time()+p.time());
		NTT::multi(&((*this)[0]),&p[0],&r[0],time(),p.time());
		return r;
	}
	inline poly operator*(const int& k)const{
		poly r(*this);
		int n=time();
		for(int i=0;i<=n;++i)r[i]=mul(r[i],k);
		return r;
	}
	inline poly operator~()const{
		poly r(*this);
		reverse(r.begin(),r.end());
		return r;
	}
};

inline poly inv(const poly& a){
	poly r(a.time());
	NTT::inver(&a[0],&r[0],a.time());
	return r;
}
inline poly der(const poly& a){
	int n=a.time();
	poly r(n-1);
	for(int i=1;i<=n;++i)r[i-1]=mul(a[i],i);
	return r;
}
int _[N];
inline poly itr(const poly& a){
	int n=a.time();
	poly r(n+1);
	_[1]=1;
	for(int i=2;i<=n+1;++i)_[i]=mul(_[mod%i],mod-mod/i);
	for(int i=0;i<=n;++i)r[i+1]=mul(a[i],_[i+1]);
	return r;
}
inline poly ln(const poly& a){
	return itr(der(a)*inv(a)%a.time());
}
inline poly exp(const poly& a){
	poly r(0,1);
	int n=a.time(),k=1;
	while(r.time()<n)
		r%=k,r=r*(a%k-ln(r)+poly(0,1))%k,k<<=1;
	return r%(n+1);
}
inline poly qpow(const poly& a,int k){
	return exp(ln(a)*k);
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    
    int i, N, M;
    cin >> N >> M;
    const int mod = 998244353;
    #define Mul mul
    #define fpow qpow
    #define Add inc
    vector<int> fac((N*M)+1), inv((N*M)+1);
    for(inv[0]=fac[0]=fac[1]=inv[1]=1,i=2;i<=(N*M);++i)
        fac[i]=Mul(fac[i-1], i), inv[i]=Mul(inv[mod%i],mod-mod/i);
    for(i=2;i<=(N*M);++i) inv[i]=Mul(inv[i-1],inv[i]);

    poly a(M*N+1);

    for(i=0;i<=M;++i) a[i]=Mul(inv[i], Mul(inv[M-i], inv[M-i]));
    int __0 = a[0], __1 = fpow(__0, mod-2);
    for(i=0;i<=M;++i) a[i]=Mul(a[i], __1);

    poly c=qpow(a, N);
    // ln(a, b, N*M+1);
    // for(i=0;i<N*M+1;++i) b[i]=Mul(b[i], N);
    // exp(b, c, N*M+1); 
    int ans=0, fMN=fpow(fac[M], N), ifMN = fpow(fMN, mod-2);
    for(i=0;i<N*M+1;++i) c[i]=Mul(c[i], ifMN);

    for(int i=0;i<=N*M;++i)
    {
        int v=Mul(fac[N*M-i], Mul(c[i], fMN));
        if(i&1) ans=Add(ans, mod-v);
        else ans=Add(ans, v);
    }
    cout << ans << "\n";

    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 13860kb

input:

2 2

output:

4

result:

ok 1 number(s): "4"

Test #2:

score: 0
Accepted
time: 0ms
memory: 13960kb

input:

5 1

output:

44

result:

ok 1 number(s): "44"

Test #3:

score: 0
Accepted
time: 24ms
memory: 14872kb

input:

167 91

output:

284830080

result:

ok 1 number(s): "284830080"

Test #4:

score: 0
Accepted
time: 0ms
memory: 13764kb

input:

2 1

output:

1

result:

ok 1 number(s): "1"

Test #5:

score: 0
Accepted
time: 2ms
memory: 14152kb

input:

2 3

output:

36

result:

ok 1 number(s): "36"

Test #6:

score: 0
Accepted
time: 2ms
memory: 13884kb

input:

2 4

output:

576

result:

ok 1 number(s): "576"

Test #7:

score: 0
Accepted
time: 0ms
memory: 13960kb

input:

3 1

output:

2

result:

ok 1 number(s): "2"

Test #8:

score: 0
Accepted
time: 0ms
memory: 14120kb

input:

3 2

output:

80

result:

ok 1 number(s): "80"

Test #9:

score: 0
Accepted
time: 0ms
memory: 13824kb

input:

3 3

output:

12096

result:

ok 1 number(s): "12096"

Test #10:

score: 0
Accepted
time: 2ms
memory: 13816kb

input:

3 4

output:

4783104

result:

ok 1 number(s): "4783104"

Test #11:

score: 0
Accepted
time: 2ms
memory: 13916kb

input:

4 1

output:

9

result:

ok 1 number(s): "9"

Test #12:

score: 0
Accepted
time: 0ms
memory: 14152kb

input:

4 2

output:

4752

result:

ok 1 number(s): "4752"

Test #13:

score: 0
Accepted
time: 2ms
memory: 13884kb

input:

4 3

output:

17927568

result:

ok 1 number(s): "17927568"

Test #14:

score: 0
Accepted
time: 3ms
memory: 13764kb

input:

4 4

output:

776703752

result:

ok 1 number(s): "776703752"

Test #15:

score: 0
Accepted
time: 0ms
memory: 13816kb

input:

5 2

output:

440192

result:

ok 1 number(s): "440192"

Test #16:

score: 0
Accepted
time: 2ms
memory: 14100kb

input:

5 3

output:

189125068

result:

ok 1 number(s): "189125068"

Test #17:

score: 0
Accepted
time: 0ms
memory: 13928kb

input:

5 4

output:

975434093

result:

ok 1 number(s): "975434093"

Test #18:

score: -100
Runtime Error

input:

1000 1000

output:


result: