Codeforces 57 C Array

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Combinatorics

Descirption

Calculate numbers of non-increasing and non-decreasing sequences with \(n\) elements ranged in \([1,n]\).

Solution

Reversing a non-decreasing array lead to a non-increasing one, thus we only need to calculate the first type.

Regarding that we choose \(i\) numbers, and increasingly use them in the array, that means split the whole array into \(i\) parts, the same as put \(i-1\) boards in \(n-1\) blanks between the elements. So the number is \(C_{n-1}^{i-1}\). And we can choose these \(i\) numbers in \(C_{n}^{i}\) ways, total numbers of non-decreasing arrays is \(\begin{align}\sum_{i = 1}^{n}{{n \choose i}{n-1 \choose i-1}}\end{align}\).

Multiply the ans by \(2\), but there‘re some re-counted situations. When all the elements are equal, it‘s both non-inc and non-dec, so subtract the ans by \(n\).

Code

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long ll;

const int maxn = 1e+5 + 5;
const ll p = 1e+9 + 7;

int n;
ll fac[maxn], inv[maxn], ans;

ll quick_power(ll base, ll index) {
  ll ret = 1;
  while (index) {
    if (index & 1) ret = ret * base % p;
    base = base * base % p;
    index >>= 1;
  }
  return ret;
}
inline ll C(int nn, int mm) {
  return fac[nn] * inv[mm] % p * inv[nn - mm] % p;
}

int main() {
  scanf("%d", &n);
  fac[0] = inv[0] = fac[1] = inv[1] = 1ll;
  for (int i = 2; i <= n; ++i) {
    fac[i] = fac[i - 1] * (ll)i % p;
    inv[i] = quick_power(fac[i], p - 2);
  }
  for (int i = 1; i <= n; ++i){
    ll tmp = C(n, i) * C(n - 1, i - 1) % p;
    ans = (ans + tmp) % p;
  }
  ans = (ans * 2ll % p - n + p) % p;
  cout << ans << endl;
  return 0;
}

Codeforces 57 C Array

标签：split   hat   ever   only   solution   blank   ati   stream   pow   

原文地址：https://www.cnblogs.com/nishikino-curtis/p/9808307.html