51nod 1228 （伯努利数 + 组合数学）

题目：

分析：

Code：

import java.util.Scanner;

public class Main {
public static Scanner cin = new Scanner(System.in);
public static int maxn = 2000 + 131;
public static long MOD = (long)(1e9 + 7);
public static long[] Inv = new long[maxn];
public static long[] B = new long[maxn];
public static long[] Mjie = new long[maxn];
public static long[] InvM = new long[maxn];

public static void INITC() {
//逆元
Inv[1] = 1;
for(int i = 2; i < maxn; ++i) {
Inv[i] = (MOD - MOD/i) * Inv[(int) (MOD % i)] % MOD;
}
// M！
Mjie[0] = Mjie[1] = 1;
for(int i = 1; i < maxn; ++i) {
Mjie[i] = Mjie[i-1] * i % MOD;
}
//阶乘逆元
InvM[1] = InvM[0] = 1;
for(int i = 2; i < maxn; ++i) {
InvM[i] = InvM[i-1] * Inv[i] % MOD;
}
}
//组合数
public static long CC(int m, int n) {
if(n == 0) return 1L;
return Mjie[m] * InvM
% MOD * InvM[m-n] % MOD;
}
//伯努利数
public static void INIT() {
B[0] = 1;
for(int i = 1; i < maxn-1; ++i) {
B[i] = 0;
for(int j = 0; j < i; ++j) {
B[i] += (CC(i+1,j) * B[j]) % MOD;
B[i] %= MOD;
}
B[i] = B[i] * Inv[i+1] * -1 % MOD;
B[i] = (B[i] + MOD) % MOD;
}
}

public static long GetAns(long N, long K) {
long Pow = 1;
long ans = 0;
for(int i = 1; i <= K+1; ++i) {
Pow = Pow * (N+1) % MOD;
ans = ans + (CC((int)(K+1),i) * B[(int)(K+1-i)]) % MOD * Pow % MOD;
ans %= MOD;
}
ans = ans * Inv[(int)(K+1)] % MOD;
return ans;
}

public static void main(String[] arg) {
INITC();
INIT();
int T;
long N, K;
T = cin.nextInt();
for(int t = 0; t < T; ++t) {
N = cin.nextLong();
K = cin.nextLong();
System.out.println(GetAns(N % MOD,K));
}
}
}