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QOJ
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 | Afterlife | AC ✓ | 272ms | 36656kb | C++20 | 5.4kb | 2024-05-22 03:12:12 | 2024-05-22 03:12:12 |
Judging History
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