CF932E

https://codeforces.com/problemset/problem/932/E

求 \(\sum\limits_{i=1}^n {n\choose i} i^k\)。

\(n\le 1e9,k\le 5e3\)

\[\begin{align*} f_{n,k} &= \sum\limits_{i=1}^n {n\choose i} i^k \\ &=\sum\limits_{i=1}^n\bigg[{n-1\choose i-1} + {n-1\choose i}\bigg]i^k \\ &=\sum\limits_{i=1}^n{n-1\choose i-1}i^k+\sum\limits_{i=1}^{n-1}{n-1\choose i}i^k \\ &=\sum\limits_{i=1}^n\frac{i}{n}{n\choose i}i^k+f_{n-1,k} \\ &=\frac{1}{n}\sum\limits_{i=1}^n{n\choose i}i^{k+1}+f_{n-1,k} \\ &=\frac{1}{n}f_{n,k+1}+f_{n-1,k} \\ 移项得 f_{n,k} = n(f_{n,k-1} - f_{n-1,k-1}) \end{align*} \]

第二维一直减，\(f_{n,0}=\sum\limits_{i=1}^n{n\choose i} =2^n-1\)。

\(n\lt k\) 时，\(f_{1,k}=1\)。

复杂度 \(O(k^2)\)。

#include <bits/stdc++.h>

using namespace std;

#define int long long

int read() {
    int s = 0, w = 1;
    char c = getchar();
    while (!isdigit(c)) {
        if (c == '-') w = -1;
        c = getchar();
    }
    while (isdigit(c)) {
        s = s * 10 + c - 48;
        c = getchar();
    }
    return s * w;
}
void pr(int x) {
	if (x < 0)
		putchar('-'), x = -x;
	if (x > 9)
		pr(x / 10);
	putchar(x % 10 + 48);
}
#define end_ putchar('\n')
#define spc_ putchar(' ')

const int maxN = 5e3 + 7;

const int mod = 1e9 + 7;

int ksm(int a, int b) {
    int res = 1;
    while (b) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

int mp[maxN][maxN], P;

int f(int n, int k) {
    if (mp[n - P][k] != -1) return mp[n - P][k];
    if (k == 0) return (ksm(2, n) - 1 + mod) % mod;
    if (n == 1) return 1;
    int res = (f(n, k - 1) - f(n - 1, k - 1) + mod) % mod * n % mod;
    mp[n - P][k] = res;
    return res;
}

signed main() {
    // freopen("1.in", "r", stdin);
    // freopen("1.out", "w", stdout);

    memset(mp, 255, sizeof(mp));

    int n = read(), k = read();
    P = n - k;
    pr(f(n, k)), end_;
}

内容来自：https://www.cnblogs.com/07Qyun/p/18462140