Pre
Write a (O (n) \) \ code as a reference.
This is something more pit.
My code to open a few among the array, becauseI'm too lazyEase of understanding.
Solution
First, we find positive solutions (Lagrange specific difference is not talked about).
Find the need to maintain:
Multiplicative inverse element factorial
Because it is from \ (1 \) start, the required space is not large, so you can \ (O (k) \) processed, the next.
\(\prod\limits_{i=1}^nx-i\)
This is not the inverse, each element can be obtained, even by re-inversion yuan, according to routine it.
\ (n \ leq k + 2 \) answers when
In order to ensure strict time complexity is \ (O (k) \) , this should be linear sieve.
Code
#include <cstdio>
using namespace std;
const int N = 1000000 + 5, mod = 1000000007;
inline int mul (int a, int b) {return 1LL * a * b % mod;}
inline int add (int a, int b) {return a + b >= mod ? a + b - mod : a + b;}
inline int mns (int a, int b) {return a - b < 0 ? a - b + mod : a - b;}
inline int qpow (int u, int v) {
int tot = 1, base = u % mod;
while (v) {
if (v & 1) tot = mul (tot, base);
base = mul (base, base);
v >>= 1;
}
return tot;
}
int x, k;
struct DuLiu {
int pri[N + 5], tot, res[N + 5];
bool vis[N + 5];
inline void init () {
int n = k + 2;
res[1] = 1;
for (int i = 2; i <= n; ++i) {
if (!vis[i]) {
pri[++tot] = i;
res[i] = qpow (i, k);
}
for (int j = 1; j <= tot; ++j) {
if (pri[j] * i > n) break;
vis[i * pri[j]] = 1;
res[i * pri[j]] = mul (res[i], res[pri[j]]);
if (i % pri[j] == 0) break;
}
}
for (int i = 2; i <= n; ++i) res[i] = add (res[i - 1], res[i]);
}
}num;
int fac[N + 5], fac_prod[N + 5], fac_prod_inv[N + 5], extra, mid[N + 5], mid_prod[N + 5], mid_prod_inv[N + 5], mid_inv[N + 5];
inline void init_mid () {
int n = k + 2;
for (int i = 1; i <= n; ++i) mid[i] = x - i; mid[0] = x;
mid_prod[0] = 1; for (int i = 1; i <= n; ++i) mid_prod[i] = mul (mid_prod[i - 1], mid[i]);
mid_prod_inv[n] = qpow (mid_prod[n], mod - 2);
for (int i = n - 1; i >= 0; --i) mid_prod_inv[i] = mul (mid_prod_inv[i + 1], mid[i + 1]);
for (int i = 1; i <= n; ++i) mid_inv[i] = mul (mid_prod_inv[i], mid_prod[i - 1]);
mid_inv[0] = qpow (x, mod - 2);
}
inline void Lagrange () {
int ans = 0, n = k + 2;
for (int i = 1; i <= n; ++i) {
int tmp = 1, f = 1;
tmp = mul (tmp, mul (fac_prod_inv[i - 1], i - 2 >= 0 ? fac_prod[i - 1 - 1] : 1));
tmp = mul (tmp, mul (fac_prod_inv[n - i], n - i - 1 >= 0 ? fac_prod[n - i - 1] : 1));
if ((n - i) % 2 == 1) f *= -1;
int liujuakioi = mid_prod[n];
tmp = mul (tmp, mul (liujuakioi, mid_inv[i]));
tmp = mul (tmp, num.res[i]);
if (f > 0) ans = add (ans, tmp);
else ans = mns (ans, tmp);
}
printf ("%d\n", ans);
}
int main () {
scanf ("%d%d", &x, &k);
num.init ();
if (x <= k + 2) {printf ("%d\n", num.res[x]); return 0;}
fac[0] = 1; for (int i = 1; i <= N; ++i) fac[i] = mul (fac[i - 1], i);
fac_prod[0] = 1; for (int i = 1; i <= N; ++i) fac_prod[i] = mul (fac_prod[i - 1], fac[i]);
fac_prod_inv[N] = qpow (fac_prod[N], mod - 2);
for (int i = N - 1; i >= 0; --i) fac_prod_inv[i] = mul (fac_prod_inv[i + 1], fac[i + 1]);
init_mid();
Lagrange ();
return 0;
}
Conclusion
More details, pay several times before too.
Note \ (Lagrange \) using a function inside a variable, error-prone (actually I wrote too much trouble, seemingly alone inverse factorial of an array of up another more convenient).