T(n) = n^k,S(n) = T(1) + T(2) + ...... T(n)。给出n和k,求S(n)。
例如k = 2,n = 5,S(n) = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55。
由于结果很大,输出S(n) Mod 1000000007的结果即可。
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
typedef long long LL;
const LL MOD = 1000000007;
const int N = 2005;
LL C[N][N];
LL B[N],Inv[N];
LL Tmp[N];
LL n;
void Init()
{
//预处理组合数
for(int i=0; i<N; i++)
{
C[i][0] = C[i][i] = 1;
if(i == 0) continue;
for(int j=1; j<i; j++)
C[i][j] = (C[i-1][j] % MOD + C[i-1][j-1] % MOD) % MOD;
}
//预处理逆元
Inv[1] = 1;
for(int i=2; i<N; i++)
Inv[i] = (MOD - MOD / i) * Inv[MOD % i] % MOD;
//预处理伯努利数
B[0] = 1;
for(int i=1; i<N; i++)
{
LL ans = 0;
if(i == N - 1) break;
for(int j=0; j<i; j++)
{
ans += C[i+1][j] * B[j];
ans %= MOD;
}
ans *= -Inv[i+1];
ans = (ans % MOD + MOD) % MOD;
B[i] = ans;
}
}
LL Work(int k)
{
LL ans = Inv[k+1];
LL sum = 0;
for(int i=1; i<=k+1; i++)
{
sum += C[k+1][i] * Tmp[i] % MOD * B[k+1-i] % MOD;
sum %= MOD;
}
ans *= sum;
ans %= MOD;
return ans;
}
int main()
{
int T;
Init();
scanf("%d", &T);
while(T--)
{
int k;
scanf("%I64d %d", &n, &k);
n %= MOD;
Tmp[0] = 1;
for(int i=1; i<N; i++)
Tmp[i] = Tmp[i-1] * (n + 1) % MOD;
printf("%I64d\n", Work(k));
}
return 0;
}
This task is very simple.
You are to calculate Σik (1 <= i <= n).
import java.math.*;
import java.util.*;
public class Main {
public static final int N = 105;
public static final BigInteger FLAG = (BigInteger.ZERO).subtract(BigInteger.ONE);
public static BigInteger[][] C = new BigInteger[N][N];
public static BigInteger[] ans = new BigInteger[N];
public static void Init(){
for(int i=0; i<N; i++){
C[i][0] = C[i][i] = BigInteger.ONE;
if(i == 0) continue;
for(int j=1; j<i; j++)
C[i][j] = C[i-1][j].add(C[i-1][j-1]);
}
}
public static BigInteger Solve(BigInteger n, int k){
if(ans[k].compareTo(FLAG) != 0){
return ans[k];
}
if(k == 1){
ans[k] = ((n.add(BigInteger.ONE)).multiply(n)).divide(BigInteger.valueOf(2));
return ans[k];
}
BigInteger tmp = BigInteger.ONE;
for(int i=0; i<k+1; i++){
tmp = tmp.multiply(n.add(BigInteger.ONE));
}
tmp = tmp.subtract(n.add(BigInteger.ONE));
BigInteger sum = BigInteger.ZERO;
for(int i=1; i<k; i++){
BigInteger t = C[k+1][i+1].multiply(Solve(n, k-i));
sum = sum.add(t);
}
ans[k] = (tmp.subtract(sum)).divide(BigInteger.valueOf(k+1));
return ans[k];
}
public static void main(String[] args){
Init();
Scanner cin = new Scanner(System.in);
while(cin.hasNext()){
BigInteger n = cin.nextBigInteger();
int k = cin.nextInt();
for(int i=0; i<N; i++){
ans[i] = FLAG;
}
System.out.println(Solve(n, k));
}
}
}