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_JUDGEfreopen("bzoj3453.in","r",stdin);freopen("bzoj3453.out","w",stdout);const char LL[]="%I64d\n"; #elseconst char LL[]="%lld\n"; #endifT=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; }