【51Nod1829】函数

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_39972971/article/details/82049798

【题目链接】

• 点击打开链接

【思路要点】

• 用容斥原理计算答案，$Ans=\sum_{i=0}^{M}(-1)^{i}*\binom{M}{i}*(M-i)^{N}$。
• 时间复杂度$O(NLogN)$。

【代码】

#include<bits/stdc++.h>

using namespace std;
const int MAXN = 1000005;
const int P = 1e9 + 7;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
  x = 0; int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
template <typename T> void write(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
  write(x);
  puts("");
}
int n, m, fac[MAXN], inv[MAXN];
int power(int x, int y) {
  if (y == 0) return 1;
  int tmp = power(x, y / 2);
  if (y % 2 == 0) return 1ll * tmp * tmp % P;
  else return 1ll * tmp * tmp % P * x % P;
}
int getc(int x, int y) {
  return 1ll * fac[x] * inv[y] % P * inv[x - y] % P;
}
int main() {
  read(n), read(m);
  fac[0] = 1;
  for (int i = 1; i <= m; i++)
      fac[i] = 1ll * fac[i - 1] * i % P;
  inv[m] = power(fac[m], P - 2);
  for (int i = m - 1; i >= 0; i--)
      inv[i] = inv[i + 1] * (i + 1ll) % P;
  int ans = 0;
  for (int i = 0; i <= m; i++) {
      int tmp = 1;
      if (i & 1) tmp = P - 1;
      ans = (ans + 1ll * tmp * getc(m, i) % P * power(m - i, n)) % P;
  }
  writeln(ans);
  return 0;
}