[SHOI2015]超能粒子炮·改

嘟嘟嘟




先看了一遍lucas，还是只能拿50分（似乎已经满足了）。

正解当然还是看某个大佬的啦。




我们要求的就是

\[f(n, k) = \sum _ {i = 0} ^ {k} C _ {n} ^ {i} \% p
\]

然后根据lucas定理，就开始愉快的推式子了……

\[\begin{align*}
f(n, k)
&= \sum _ {i = 0} ^ {k} C _ {n} ^ {i} \% p \\
&= \sum _ {i = 0} ^ {k} C_ {n / p} ^ {i / p} * C _ {n \% p} ^ {i \% p} \\
&= C_{n / p} ^ {0} \sum _ {i = 0} ^ {p - 1} C _ {n \% p} ^ {i} + C _ {n / p} ^ {1} \sum _ {i = 0} ^ {p - 1} C _ {n \% p} ^ {i} + \ldots + C_{n / p} ^ {k / p - 1} \sum _ {i = 0} ^ {p - 1} C _ {n \% p} ^ {i} + C _ {n / p} ^ {k / p} \sum _ {i = 0} ^ {k \% p} C _ {n \% p} ^ {i}\\
&= \sum _ {i = 0} ^ {p - 1} C _ {n \% p} ^ {i} * (C _ {n / p} ^ {0} + C_{n / p} ^ {1} + \ldots + C_{n / p} ^ {k / p - 1}) + C _ {n / p} ^ {k / p} \sum _ {i = 0} ^ {k \% p} C _ {n \% p} ^ {i} \\
&= f(n \% p, p - 1) * f(n / p, k / p - 1) + C _ {n / p} ^ {k / p} * f(n \% p, k \% p) \\
\end{align*}
\]

然后把\(f(n \% p, k \% p)\)预处理一下，就完事了。（说的真轻松……）

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2505;
const int p = 2333;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

ll n, K;
ll C[maxn][maxn], f[maxn][maxn];

In ll inc(ll a, ll b) {return a + b >= p ? a + b - p : a + b;}

In ll lucas(ll n, ll m)
{
  if(!m || n == m) return 1;
  if(n < m) return 0;
  return C[n % p][m % p] * lucas(n / p, m / p) % p;
}
In ll F(ll n, ll K)
{
  if(K < 0) return 0;
  if(!n || !K) return 1;
  if(n < p && K < p) return f[n][K];
  return inc(f[n % p][p - 1] * F(n / p, K / p - 1) % p, lucas(n / p, K / p) * f[n % p][K % p] % p);
}

In void init()
{
  for(int i = 0; i < maxn; ++i) C[i][0] = 1;
  for(int i = 1; i < maxn; ++i)
    for(int j = 1; j <= i; ++j)
      C[i][j] = inc(C[i - 1][j - 1], C[i - 1][j]);
  for(int i = 0; i < maxn; ++i) f[i][0] = 1;
  for(int i = 0; i < maxn; ++i)
    for(int j = 1; j < maxn; ++j)
      f[i][j] = inc(C[i][j], f[i][j - 1]);
}

int main()
{
  init();
  int T = read();
  while(T--)
    {
      n = read(), K = read();
      write(F(n, K)), enter;
    }
  return 0;
}