loj #6247. 九個太陽 k次單位根 神仙構造 FFT求和原理
阿新 • • 發佈:2020-07-30
LINK:九個太陽
不可做系列.
構造比較神仙.
考慮FFT的求和原理有 \(\frac{1}{k}\sum_{j=0}^{k-1}(w_k^j)^n=[k|n]\)
帶入這道題的式子.
有\(\sum_{i=0}^n\frac{1}{k}\sum_{j=0}^{k-1}(w_k^j)^iC(n,i)\)
顛倒求和符號 二項式定理合並即可klogn求.
k次單位根在mod 998244353時就是 \(\frac{mod-1}{k}\)
code
//#include<bits/stdc++.h> #include<iostream> #include<cstdio> #include<ctime> #include<cctype> #include<queue> #include<deque> #include<stack> #include<iostream> #include<iomanip> #include<cstdio> #include<cstring> #include<string> #include<ctime> #include<cmath> #include<cctype> #include<cstdlib> #include<queue> #include<deque> #include<stack> #include<vector> #include<algorithm> #include<utility> #include<bitset> #include<set> #include<map> #define ll long long #define db double #define INF 1000000000 #define inf 100000000000000000ll #define ldb long double #define pb push_back #define put_(x) printf("%d ",x); #define get(x) x=read() #define gt(x) scanf("%d",&x) #define gi(x) scanf("%lf",&x) #define put(x) printf("%d\n",x) #define putl(x) printf("%lld\n",x) #define rep(p,n,i) for(RE ll i=p;i<=n;++i) #define go(x) for(ll i=lin[x],tn=ver[i];i;tn=ver[i=nex[i]]) #define fep(n,p,i) for(RE ll i=n;i>=p;--i) #define vep(p,n,i) for(RE ll i=p;i<n;++i) #define pii pair<ll,ll> #define mk make_pair #define RE register #define P 1000000007ll #define gf(x) scanf("%lf",&x) #define pf(x) ((x)*(x)) #define uint unsigned long long #define ui unsigned #define EPS 1e-10 #define sq sqrt #define S second #define F first #define mod 998244353 #define id(i,j) ((i-1)*m+j) #define max(x,y) ((x)<(y)?y:x) #define a(i) t[i].a #define b(i) t[i].b using namespace std; char *fs,*ft,buf[1<<15]; inline char gc() { return (fs==ft&&(ft=(fs=buf)+fread(buf,1,1<<15,stdin),fs==ft))?0:*fs++; } inline ll read() { RE ll x=0,f=1;RE char ch=gc(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=gc();} return x*f; } //我心裡住著一位天使 我怎能可以讓她沒有翅膀? const ll G=3; ll n,k; inline ll ksm(ll b,ll p) { ll cnt=1;p=p%(mod-1); while(p) { if(p&1)cnt=(ll)cnt*b%mod; p=p>>1;b=(ll)b*b%mod; } return cnt; } signed main() { freopen("1.in","r",stdin); get(n);get(k); ll ans=0; ll wn=ksm(G,(mod-1)/k),cc=1; rep(0,k-1,i) { ans=(ans+ksm(cc+1,n))%mod; cc=(ll)cc*wn%mod; } put(ans*(ll)ksm(k,mod-2)%mod); return 0; }