F. The Sum of the k-th Powers

题目连接：

http://www.codeforces.com/contest/622/problem/F

Description

There are well-known formulas: , , . Also mathematicians found similar formulas for higher degrees.

Find the value of the sum modulo 109 + 7 (so you should find the remainder after dividing the answer by the value 109 + 7).

Input

The only line contains two integers n, k (1 ≤ n ≤ 109, 0 ≤ k ≤ 106).

Output

Print the only integer a — the remainder after dividing the value of the sum by the value 109 + 7.

Sample Input

4 1

Sample Output

Hint

题意

让你计算1^k+2k+....+n^k

题解：

拉格朗日插值法

答案等于$${P}{x} = \sum{i}^{k+2}({P}{i}\prod{j=1,j\neq i}^{k+2}\frac{n-j}{i-j})$$

最后的答案就等于P(n)

我们预处理(n-j)的阶乘，再预处理下面的阶乘就好了

对于这样，对于每一个i，我们都能够O(logn)来计算了(logn拿来求逆元)

代码

#include<bits/stdc++.h>

using namespace std;

const int mod = 1e9+7;

const int maxn = 1e6+7;

long long quickpow(long long  m,long long n,long long k)//返回m^n%k

{

    long long b = 1;

    while (n > 0)

    {

          if (n & 1)

             b = (b*m)%k;

          n = n >> 1 ;

          m = (m*m)%k;

    }

    return b;

}

long long p[maxn];

long long fac[maxn];

int n,k;

int main()

{

    fac[0]=1;

    for(int i=1;i<maxn;i++)

        fac[i]=(fac[i-1]*i)%mod;

    scanf("%d%d",&n,&k);

    p[0]=0;

    for(int i=1;i<=k+2;i++)

        p[i]=(p[i-1]+quickpow(i,k,mod))%mod;

    if(n<=k+2)

    {

        printf("%d\n",p[n]);

        return 0;

    }

    long long cur = 1;

    for(int i=1;i<=k+2;i++)

        cur=(cur*(n-i))%mod;

    long long ans = 0;

    for(int i=1;i<=k+2;i++)

    {

        long long tmp = quickpow(fac[k+2-i]%mod*fac[i-1]%mod,mod-2,mod);

        long long tmp2 = quickpow(n-i,mod-2,mod);

        if((k+2-i)%2)tmp=-tmp;

        ans =(ans + p[i]*cur%mod*tmp%mod*tmp2)%mod;

    }

    cout<<ans<<endl;

}

巴特西

Educational Codeforces Round 7 F. The Sum of the k-th Powers 拉格朗日插值法