51nod 1258序列求和

T(n) = n^k，S(n) = T(1) + T(2) + … T(n)。給出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。
由於結果很大，輸出S(n) Mod 1000000007的結果即可。

拉格朗日插值：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
const int maxn = 50000 + 10;
const int N = maxn;
ll fac[maxn], pl[maxn], pr[maxn];
ll qpow(ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) {
			ans = (ans * a) % mod;
		}
		a = (a * a) % mod;
		b >>= 1;
	}
	return ans % mod;
}
int main() {
	 fac[0] = 1;
	for (int i = 1; i <= N; i++) {
		fac[i] = fac[i - 1] * i % mod;
	}
	int T;
	scanf("%d", &T);
	while (T--) {
		ll n, k, y=0, ans = 0;
		scanf("%lld%lld", &n, &k);
		pl[0] = pr[k + 3] = 1;
		for (int i = 1; i <= k + 2; i++) {
			pl[i] = pl[i - 1] * (n % mod - i + mod) % mod;
		}
		for (int i = k + 2; i >= 1; i--) {
			pr[i] = pr[i + 1] * (n % mod - i + mod) % mod;
		}
		for (int i = 1; i <= k + 2; i++) {
			y = (y + qpow(i, k)) % mod;
			ll a = pl[i - 1] * pr[i + 1] % mod;
			ll b = fac[i - 1] * ((k + 2 - i) & 1 ? -1 : 1) * fac[k + 2 - i];
			b = (b % mod + mod) % mod;
			ans = (ans + y * a % mod * qpow(b, mod - 2) % mod)  % mod;
		}
		printf("%lld\n", ans);
	}
	return 0;
}