題目鏈接:點擊查看
題目大意:中文題意,不難理解
題目分析:正解是需要給原函數求導,用矩陣快速冪求出導函數差分後的值,再推出原函數的值(我是沒看懂,畢竟是防ak的題),但後來lb學長來問我能不能用BM做,我一臉疑惑:什麼是BM?去學了一波黑科技後,發現可以用BM導入前500項然後水過去。。我就說爲什麼這個題比賽的時候過的人數那麼多
代碼:
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <cassert>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=998244353;
ll powmod(ll a,ll b)
{
ll res=1;
a%=mod;
assert(b>=0);
for(;b;b>>=1)
{
if(b&1)
res=res*a%mod;
a=a*a%mod;
}
return res;
}
int _,n;
namespace linear_seq
{
const int N=10010;
ll res[N],base[N],_c[N],_md[N];
vector<int> Md;
void mul(ll *a,ll *b,int k) {
for(int i = 0 ; i < k + k ; ++i)
_c[i]=0;
for(int i = 0 ; i < k ;++i)
if (a[i])
for(int j = 0 ;j < k ;++ j)
_c[i+j]=(_c[i+j]+a[i]*b[j])%mod;
for (int i=k+k-1;i>=k;i--)
if (_c[i])
for(int j = 0 ; j<(int ) Md.size() ; ++ j)
_c[i-k+Md[j]]=(_c[i-k+Md[j]]-_c[i]*_md[Md[j]])%mod;
for(int i =0 ; i< k ; ++i)
a[i]=_c[i];
}
int solve(ll n,VI a,VI b) {
ll ans=0,pnt=0;
int k=SZ(a);
assert( SZ(a) == SZ(b) );
for(int i = 0 ;i < k ; ++ i)
_md[k-1-i] = -a[i] ; _md[k] = 1 ;
Md.clear() ;
for(int i =0 ; i < k ; ++ i)
if (_md[i]!=0)
Md.push_back(i);
for(int i = 0; i< k ;++ i)
res[i]=base[i]=0;
res[0]=1;
while ((1ll<<pnt)<=n)
pnt++;
for (int p=pnt;p>=0;p--) {
mul(res,res,k);
if ((n>>p)&1) {
for (int i=k-1;i>=0;i--) res[i+1]=res[i];res[0]=0;
for(int j = 0 ;j < (int)Md.size() ; ++ j)
res[ Md[j] ]=(res[ Md[j] ]-res[k]*_md[Md[j]])%mod;
}
}
rep(i,0,k) ans=(ans+res[i]*b[i])%mod;
if (ans<0) ans+=mod;
return ans;
}
VI BM(VI s) {
VI C(1,1),B(1,1);
int L=0,m=1,b=1;
for(int n= 0 ;n < (int)s.size(); ++ n ) {
ll d=0;
for(int i =0 ; i < L +1 ;++ i)
d=(d+(ll)C[i]*s[n-i])%mod;
if (d==0) ++m;
else if (2*L<=n) {
VI T=C;
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)<SZ(B)+m)
C.push_back(0);
for(int i =0 ; i < (int)B.size(); ++ i)
C[i+m]=(C[i+m]+c*B[i])%mod;
L=n+1-L; B=T; b=d; m=1;
} else {
ll c=mod-d*powmod(b,mod-2)%mod;
while (SZ(C)<SZ(B)+m)
C.push_back(0);
for(int i = 0 ;i <(int) B.size() ; ++ i)
C[i+m]=(C[i+m]+c*B[i])%mod;
++m;
}
}
return C;
}
ll gao(VI a,ll n) {
VI c=BM(a);
c.erase(c.begin());
for( int i = 0 ; i < (int)c.size( );++i )
c[i]=(mod-c[i])%mod;
return (ll)solve(n,c,VI(a.begin(),a.begin()+SZ(c)));
}
};
ll fib[510];
void init()
{
fib[1]=1;
fib[2]=1;
for(int i=3;i<=500;i++)
fib[i]=(fib[i-1]+fib[i-2])%mod;
}
int main() {
init();
VI a;
ll n,k;
scanf("%lld%lld",&n,&k);
ll ans=0;
for(int i=1;i<=500;i++)
{
ans=(ans+powmod(i,k)*fib[i]%mod)%mod;
a.push_back(ans);
}
printf("%lld\n",linear_seq::gao(a,n-1));
return 0 ;
}