51nod1228

伯努利数

这个是答案

其中的b是伯努利数，可以n^2预处理

伯努利数n^2递推

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e3 + , mod = 1e9 + ;
ll n, k;
ll inv[N], c[N][N], b[N];
inline ll rd()
{
    ll x = , f = ;
    char c = getchar();
    while(c < '' || c > '') { if(c == '-') f = -; c = getchar(); }
    while(c >= '' && c <= '') { x = x *  + c - ''; c = getchar(); }
    return x * f;
}
int main()
{
    int T = rd();
    c[][] = ;
    for(int i = ; i < N; ++i)
    {
        c[i][] = ;
        for(int j = ; j < N; ++j) c[i][j] = (c[i - ][j] + c[i - ][j - ]) % mod;
    }
    inv[] = ;
    for(int i = ; i < N; ++i)
        if(i != ) inv[i] = (mod - mod / i) * inv[mod % i] % mod;
    b[] = ;
    for(int i = ; i < N - ; ++i)
    {
        for(int j = ; j < i; ++j)
            b[i] = (b[i] + c[i + ][j] * b[j]) % mod;
        b[i] = ((b[i] * -inv[i + ] % mod) + mod) % mod;
    }
    while(T--)
    {
        n = rd() % mod;
        k = rd();
        ll ans = , fac = ;
        for(int i = ; i <= k + ; ++i)
        {
            fac = fac * (n + ) % mod;
            ans = (ans + c[k + ][i] * b[k +  - i] % mod * fac % mod) % mod;
        }
        ans = (ans * inv[k + ]) % mod;
        printf("%lld\n", ans);
    }
    return ;
}