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 + 5, mod = 1e9 + 7;
ll n, k;
ll inv[N], c[N][N], b[N];
inline ll rd()
{
    ll 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 * 10 + c - '0'; c = getchar(); }
    return x * f;
}
int main()
{
    int T = rd();
    c[0][0] = 1;
    for(int i = 1; i < N; ++i)
    {
        c[i][0] = 1;
        for(int j = 1; j < N; ++j) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
    inv[1] = 1;
    for(int i = 1; i < N; ++i)
        if(i != 1) inv[i] = (mod - mod / i) * inv[mod % i] % mod;
    b[0] = 1;
    for(int i = 1; i < N - 1; ++i)
    {
        for(int j = 0; j < i; ++j) 
            b[i] = (b[i] + c[i + 1][j] * b[j]) % mod;
        b[i] = ((b[i] * -inv[i + 1] % mod) + mod) % mod;
    }
    while(T--)
    {
        n = rd() % mod;
        k = rd();
        ll ans = 0, fac = 1;
        for(int i = 1; i <= k + 1; ++i) 
        {
            fac = fac * (n + 1) % mod;
            ans = (ans + c[k + 1][i] * b[k + 1 - i] % mod * fac % mod) % mod;
        }    
        ans = (ans * inv[k + 1]) % mod;
        printf("%lld
", ans);
    }
    return 0;
}