uoj#269. 【清華集訓2016】如何優雅地求和(數論)
首先,如果\(f(x)=1\),那麼根據二項式定理,有\(Q(f,n,k)=1\)
當\(f(x)=x\)的時候,有\[Q=\sum_{i=0}^ni\times \frac{n!}{i!(n-i)!}k^i(1-k)^{n-i}\]
\[Q=\sum_{i=0}^nnk\times \frac{(n-1)!}{(i-1)!(n-i)!}k^{i-1}(1-k)^{n-i}\]
\[Q=nk\sum_{i=0}^n\frac{(n-1)!}{(i-1)!(n-i)!}k^{i-1}(1-k)^{n-i}\]
\[Q=nk\sum_{i=0}^n{n-1\choose i-1}k^{i-1}(1-k)^{n-i}\]
根據二項式定理後面的等於\(1\),所以\(Q=nk\)
然後我們發現,如果\(f(x)=x^{\underline{d}}\),則有\(Q=n^{\underline{d}}k^d\),其中\(x^{\underline{d}}\)是\(x\)的\(d\)次下降冪,為\(x(x-1)...(x-d+1)\),證明和上面的差不多當
\[Q=\sum_{i=0}^ni^{\underline{d}}\times \frac{n!}{i!(n-i)!}k^i(1-k)^{n-i}\]
\[Q=\sum_{i=0}^nn^{\underline{d}}x^d\times \frac{(n-d)!}{(i-d)!(n-i)!}k^{i-d}(1-k)^{n-i}\]
\[Q=n^{\underline{d}}k^d\sum_{i=0}^n \frac{(n-d)!}{(i-d)!(n-i)!}k^{i-d}(1-k)^{n-i}\]
\[Q=n^{\underline{d}}k^d\sum_{i=0}^n{n-d\choose i-d}k^{i-d}(1-k)^{n-i}\]
後面那個還是等於\(1\)
根據乘法分配律,如果\(f(x)=\sum_{i=0}^m a_ix^{\underline{i}}\),那麼\(Q(f,n,x)=\sum_{i=0}^m a_i\times Q(x^{\underline{i}},n,k)\)
考慮如何計算\(a_i\)
點值轉系數來求出\(b_i\),複雜度\(O(m\log m)\)
然而\(FFT\)太煩了而且std寫的是\(O(m^2)\)的我們不能辜負出題人的一片好心,所以來考慮\(O(m^2)\)暴力
當\(x=0\)時,\(f(x)=b_0\)
記\(\triangle f(x)=f(x+1)-f(x)\),即一階差分,因為\({x+1\choose i}-{x\choose i}={x\choose i-1}\),所有\(\triangle f(x)=\sum_{i=1}^m b_i{x\choose i-1}\),那麼\(\triangle f(0)=b_1\)
同理,\(\triangle^kf(0)=b_k\),即\(k\)階差分後\(0\)處的值為\(b_k\)
然後沒有然後了
//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=a,I=b+1;i<I;++i)
#define fd(i,a,b) for(R int i=a,I=b-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
R int res,f=1;R char ch;
while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
return res*f;
}
const int N=2e4+5,P=998244353;
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
int ksm(R int x,R int y){
R int res=1;
for(;y;y>>=1,x=mul(x,x))if(y&1)res=mul(res,x);
return res;
}
int b[N],n,m,k,ans,p=1;
int main(){
// freopen("testdata.in","r",stdin);
n=read(),m=read(),k=read();
fp(i,0,m)b[i]=read();
fp(i,0,m){
ans=add(ans,mul(p,b[0]));
fp(j,0,m-i-1)b[j]=dec(b[j+1],b[j]);
p=1ll*p*k%P*(n-i)%P*ksm(i+1,P-2)%P;
}printf("%d\n",ans);
return 0;
}