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<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int T,k,a,n,d,v[N],f[N];
int ksm(int a,int k)
{
    int s=1;
    for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
    return s;
}
int inv(int a){return ksm(a,P-2);}
int calc(int n,int x)
{
    int ans=0;
    for (int i=0;i<n;i++)
    {
        int u=1;for (int j=0;j<n;j++) if (i!=j) u=1ll*u*(P+i-j)%P;
        u=1ll*v[i]*inv(u)%P;
        for (int j=0;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
";
#else
    const char LL[]="%lld
";
#endif
    T=read();
    while (T--)
    {
        k=read(),a=read(),n=read(),d=read();
        memset(v,0,sizeof(v));
        for (int i=1;i<=k+2;i++) v[i]=(v[i-1]+ksm(i,k))%P;
        for (int i=1;i<=k+2;i++) v[i]=(v[i]+v[i-1])%P;
        for (int i=0;i<=k+3;i++) f[i]=calc(k+3,(a+1ll*i*d)%P);
        for (int i=1;i<=k+3;i++) f[i]=(f[i]+f[i-1])%P;
        memcpy(v,f,sizeof(v));cout<<calc(k+4,n)<<endl;
    }
    return 0;
}