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

Posted 2022-04-11 dillonh

题目链接

传送门

题面

题意

给你\\(n,k\\)，要你求\\(\\sum\\limits_i=1^ni^k\\)的值。

思路

根据数学知识或者说题目提示可知\\(\\sum\\limits_i=1^ni^k\\)可以被一个\\(k+1\\)次多项式表示。
由拉格朗日插值法(推荐学习博客)的公式：\\(L(x)=l(x)\\sum\\limits_i=1^k+2y_i\\fracw_ix-x_i,\\text其中l(x)=\\prod\\limits_i=1^k+2(x-i),y_i=\\sum\\limits_j=1^ij^k,w_i=\\prod\\limits_j=1,j\\not= i^n\\frac1x_i-x_j\\)可以得到结果。
由于本题的特殊性，可以将\\(w_i\\)进行化简：
\\[ \\beginaligned w_i&=\\prod\\limits_j=1,j\\not= i^n\\frac1x_i-x_j&\\&=\\prod\\limits_j=1,j\\not= i^n\\frac1i-j&\\&=\\frac1(i-1)(i-2)*\\dots*1*(i-(i+1))\\dots(i-(k+2))&\\&=(-1)^k+2-i\\frac1(i-1)!(k+2-i)!& \\endaligned \\]
因此我们可以通过\\(O(k+2)\\)的复杂度得到\\(l(x),y_i,x-x_i\\)，然后通过预处理阶乘的逆元我们可以\\(O((k+2)log(k+2))\\)得到\\(w_i\\)，所以总复杂度为在\\(O((k+2)log(k+2)+(k+2))\\)左右。

代码实现如下

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 1e6 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

int n, k, pp;
int A[maxn], y[maxn], inv[maxn], w[maxn];

int qpow(int x, int n) 
    int res = 1;
    while(n) 
        if(n & 1) res = 1LL * res * x % mod;
        x = 1LL * x * x % mod;
        n >>= 1;
    
    return res;


void init() 
    A[0] = pp = 1;
    for(int i = 1; i <= min(n, k + 2); ++i) 
        A[i] = 1LL * A[i-1] * i % mod;
        inv[i] = qpow(n - i, mod - 2);
        pp = (1LL * pp * (n - i) % mod + mod) % mod;
        y[i] = (y[i-1] + qpow(i, k)) % mod;
    
    for(int i = 1; i <= min(n, k + 2); ++i) 
        w[i] = 1LL * A[i-1] * A[k+2-i] % mod;
        if((k + 2 - i) & 1) w[i] = mod - w[i];
        w[i] = qpow(w[i], mod - 2);
    


int main() 
    scanf("%d%d", &n, &k);
    init();
    if(n <= k + 2) return printf("%d\\n", y[n]) * 0;
    int ans = 0;
    for(int i = 1; i <= (k + 2); ++i) 
        ans = (ans + 1LL * pp * y[i] % mod * w[i] % mod * inv[i] % mod) % mod;
    
    printf("%d\\n", ans);
    return 0;