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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
#416694#8715. 放苹果AfterlifeAC ✓272ms36656kbC++205.4kb2024-05-22 03:12:122024-05-22 03:12:12

Judging History

This is the latest submission verdict.

  • [2024-05-22 03:12:12]
  • Judged
  • Verdict: AC
  • Time: 272ms
  • Memory: 36656kb
  • [2024-05-22 03:12:12]
  • Submitted

answer

#include<bits/stdc++.h>
using namespace std;
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);
}

inline poly operator/(const poly& f,const poly& g){
	int n=f.time()-g.time()+1;
	return ~((~f)*inv((~g)%n)%n);
}
inline poly operator%(const poly& f,const poly& g){
	if(f.time()<g.time())return f;
	return (f-g*(f/g)%g.time())%g.time();
}

int t[200005];
int rt[200005];
int C(int a,int b) {
    return 1LL * t[a] * rt[b] % mod * rt[a - b] % mod;
}
int main() {
    int n , m ; cin >> n >> m;
    t[0] = 1; rt[0] = 1;
    for(int i = 1;i <= n + 1;i++) {
        t[i] = 1LL * t[i - 1] * i % mod; rt[i] = qpow(t[i] , mod - 2);
    }

    poly g1(n + 1) , g2(n + 1) ;
    for(int i = 0;i <= n;i++) {
         g2[i] = rt[i + 1];
        g1[i] = 1LL * qpow(m , i + 1) * rt[i + 1] % mod;
        g1[i] = (g1[i] - g2[i] + mod) % mod;
    }
    poly g = g1 * inv(g2);
    for(int i = 0;i <= n;i++) {
        g[i] = 1LL * t[i ] * g[i] % mod;
    }
    poly a(n + 1);
    for(int i = 0;i <= n;i++) {
        a[i] = rt[i];
        if(i & 1) a[i] = (mod - a[i]) % mod;
    }
    poly b(n + 1);
    for(int i = 0;i <= n;i++) {
        b[i] = 1LL * rt[i] * qpow(m , i) % mod * g[n - i] % mod;
        // printf("%d %d %d\n",i,a[i],b[i]);
    }
    poly f = a * b; f.resize(n + 1);
    int ans = 0;
    for(int i = 0;i <= n;i++) {
        f[i] = 1LL * f[i] * t[i] % mod;
        // printf("%d %d\n",i,f[i]);
    }
    reverse(f.begin() , f.end()) ;
    for(int i = 1;i <= n;i++) ans = (ans + 1LL * f[i] * min(i , n - i) % mod * C(n , i)) % mod;
    cout << ans << '\n' ;
}

这程序好像有点Bug,我给组数据试试?

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 16076kb

input:

2 3

output:

8

result:

ok 1 number(s): "8"

Test #2:

score: 0
Accepted
time: 0ms
memory: 16080kb

input:

3 3

output:

36

result:

ok 1 number(s): "36"

Test #3:

score: 0
Accepted
time: 2ms
memory: 15732kb

input:

1 1

output:

0

result:

ok 1 number(s): "0"

Test #4:

score: 0
Accepted
time: 2ms
memory: 14032kb

input:

1 2

output:

0

result:

ok 1 number(s): "0"

Test #5:

score: 0
Accepted
time: 0ms
memory: 13720kb

input:

1 3

output:

0

result:

ok 1 number(s): "0"

Test #6:

score: 0
Accepted
time: 2ms
memory: 13732kb

input:

2 1

output:

0

result:

ok 1 number(s): "0"

Test #7:

score: 0
Accepted
time: 3ms
memory: 15820kb

input:

3 1

output:

0

result:

ok 1 number(s): "0"

Test #8:

score: 0
Accepted
time: 2ms
memory: 13968kb

input:

3719 101

output:

78994090

result:

ok 1 number(s): "78994090"

Test #9:

score: 0
Accepted
time: 0ms
memory: 13896kb

input:

2189 1022

output:

149789741

result:

ok 1 number(s): "149789741"

Test #10:

score: 0
Accepted
time: 4ms
memory: 14184kb

input:

2910 382012013

output:

926541722

result:

ok 1 number(s): "926541722"

Test #11:

score: 0
Accepted
time: 241ms
memory: 32560kb

input:

131072 3837829

output:

487765455

result:

ok 1 number(s): "487765455"

Test #12:

score: 0
Accepted
time: 261ms
memory: 34316kb

input:

183092 100000000

output:

231786691

result:

ok 1 number(s): "231786691"

Test #13:

score: 0
Accepted
time: 272ms
memory: 35356kb

input:

197291 937201572

output:

337054675

result:

ok 1 number(s): "337054675"

Test #14:

score: 0
Accepted
time: 265ms
memory: 34576kb

input:

200000 328194672

output:

420979346

result:

ok 1 number(s): "420979346"

Test #15:

score: 0
Accepted
time: 267ms
memory: 36656kb

input:

200000 1000000000

output:

961552572

result:

ok 1 number(s): "961552572"

Extra Test:

score: 0
Extra Test Passed