神犇LYD虐完HEOI之后给傻×XLk出了一题：
SHY是某国的公主，平时的一大爱好是读诗...(中间略)...结果mod p就可以了

简明题意
给定 k,a,n,d,p
f(i)=1^k+2^k+3^k+......+i^k
g(x)=f(1)+f(2)+f(3)+....+f(x)
求(g(a)+g(a+d)+g(a+2d)+......+g(a+nd))mod p

对于所有数据
1<=k<=123
0<=a,n,d<=123456789
p==1234567891

第一行数据组数，(保证小于6)
以下每行四个整数 k,a,n,d

每行一个结果。

5
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1

5
5
5
5
5

 1 #include<cstring>
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<algorithm>
 5 #include<cmath>
 6
 7 #define p 1234567891
 8 #define N 157
 9 #define ll long long
10 using namespace std;
11 inline int read()
12 {
13     int x=0,f=1;char ch=getchar();
14     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
15     while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
16     return x*f;
17 }
18
19 ll a,n,d,m,k;
20 ll s1[N],s2[N];
21 ll g[N],f[N],inv[N<<1];
22
23 ll fast_pow(ll a,ll b)
24 {
25     ll ans=1;
26     while(b)
27     {
28         if (b&1) (ans*=a)%=p;
29         (a*=a)%=p;
30         b>>=1;
31     }
32     return ans;
33 }
34 inline ll Lagrange(ll *a,int n,ll pos)
35 {
36     if (pos<=n) return a[pos];
37     ll ans=0;
38     for (int i=1;i<=n;i++)
39     {
40         ll s1=1,s2=1;
41         for (int j=1;j<=n;j++)
42             if (i!=j)
43             {
44                 (s1*=(pos-j))%=p;
45                 (s2*=(i-j))%=p;
46             }
47         (ans+=a[i]*s1%p*fast_pow(s2,p-2))%=p;
48     }
49     return ans;
50 }
51 int main()
52 {
53     freopen("fzy.in","r",stdin);
54     freopen("fzy.out","w",stdout);
55
56     int T=read();
57     while(T--)
58     {
59         k=read(),a=read(),n=read(),d=read();
60         for (int i=1;i<=k+3;i++) g[i]=fast_pow(i,k);
61         for (int i=2;i<=k+3;i++) (g[i]+=g[i-1])%=p;
62         for (int i=2;i<=k+3;i++) (g[i]+=g[i-1])%=p;
63         f[0]=Lagrange(g,k+3,a);
64         for (int i=1;i<=k+5;i++) f[i]=Lagrange(g,k+3,(i*d+a)%p),(f[i]+=f[i-1])%=p;
65         printf("%lld\n",(Lagrange(f,k+5,n)+p)%p);
66     }
67 }