BZOJ3453 XLkxc(拉格朗日插值)

  显然f(i)是一个k+2项式,g(x)是f(i)的前缀和,则显然其是k+3项式,插值即可。最后要求的东西大胆猜想是个k+4项式继续插值就做完了。注意2p>maxint……

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define int long long
#define P 1234567891
#define N 200
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
int x=,f=;char c=getchar();
while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
return x*f;
}
int T,k,a,n,d,v[N],f[N];
int ksm(int a,int k)
{
int s=;
for (;k;k>>=,a=1ll*a*a%P) if (k&) s=1ll*s*a%P;
return s;
}
int inv(int a){return ksm(a,P-);}
int calc(int n,int x)
{
int ans=;
for (int i=;i<n;i++)
{
int u=;for (int j=;j<n;j++) if (i!=j) u=1ll*u*(P+i-j)%P;
u=1ll*v[i]*inv(u)%P;
for (int j=;j<n;j++) if (i!=j) u=1ll*u*(P+x-j)%P;
ans=(ans+u)%P;
}
return ans;
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("bzoj3453.in","r",stdin);
freopen("bzoj3453.out","w",stdout);
const char LL[]="%I64d\n";
#else
const char LL[]="%lld\n";
#endif
T=read();
while (T--)
{
k=read(),a=read(),n=read(),d=read();
memset(v,,sizeof(v));
for (int i=;i<=k+;i++) v[i]=(v[i-]+ksm(i,k))%P;
for (int i=;i<=k+;i++) v[i]=(v[i]+v[i-])%P;
for (int i=;i<=k+;i++) f[i]=calc(k+,(a+1ll*i*d)%P);
for (int i=;i<=k+;i++) f[i]=(f[i]+f[i-])%P;
memcpy(v,f,sizeof(v));cout<<calc(k+,n)<<endl;
}
return ;
}
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