【SHOI2015】超能粒子炮·改
Description
给定n，k，求 $(\sum\limits_{i=0}^{k}{C_n^i})\ mod\ 2333$
Solution
首先，我们要用到卢卡斯定理：$$C_n^m\ mod\ p=C_{n/p}^{m/p}*C_{n\ mod\ p}^{m\ mod\ p}\ mod\ p$$
我们要求解的是$$(\sum\limits_{i=0}^{k}{C_n^i})\ mod\ 2333$$
我们定义$f(n,k)=\sum\limits_{i=0}^{k}{C_n^i}$
由卢卡斯定理可知$$\sum\limits_{i=0}^{k}{C_n^i}=\sum\limits_{i=0}^{k}{C_{n/p}^{i/p}*C_{n\ mod\ p}^{i\ mod\ p}}$$
根据整除分块（数论分块）,可得
$$=C_{n/p}^0\sum\limits_{i=0}^{p-1}{C_{n\ mod\ p}^i} + C_{n/p}^1\sum\limits_{i=0}^{p-1}{C_{n\ mod\ p}^i}+C_{n/p}^2\sum\limits_{i=0}^{p-1}{C_{n\ mod\ p}^i}+...+C_{n/p}^{k/p}\sum\limits_{i=0}^{k\ mod\ p}{C_{n\ mod\ p}^i}$$
提取公因式，得
$$=\sum\limits_{i=0}^{p-1}{C_{n\ mod\ p}^i}*({C_{n/p}^0+C_{n/p}^1+...+C_{n/p}^{k/{p-1}}})=f(n\ mod\ p, p-1)*f(n/p,k/p-1)$$
加上最后不完整的一块，得
$$f(n,k)=f(n\ mod\ p, p-1)*f(n/p,k/p-1)+C_{n/p}^{k/p}*f(n\ mod\ p, k\ mod\ p)$$
取模之后的值肯定小于2333，我们可以直接预处理[0,2333]之间的C和f,组合数使用Lucas计算，时间复杂度为$O(2333^2+Tlog_{2333}^2n)$
Code
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int mod = 2333; 5 ll C[2340][2340], f[2340][2340]; 6 ll Locas(ll n, ll m) { 7 if (m == 0) return 1; 8 if (n == m) return 1; 9 if (n < m) return 0; 10 return Locas(n / mod, m / mod) * C[n % mod][m % mod] % mod; 11 } 12 ll calc(ll x, ll y) { 13 if (y < 0) return 0; 14 if (!x || !y) return 1; 15 if (x < mod && y < mod) return f[x][y]; 16 return (calc(x / mod, y / mod - 1) * f[x % mod][mod - 1] % mod + Locas(x / mod, y / mod) * f[x % mod][y % mod] % mod) % mod; 17 } 18 int main() { 19 for (register int i = 0; i <= 2335; ++i) C[i][i] = C[i][0] = 1; 20 for (register int i = 1; i <= 2335; ++i) 21 for (register int j = 1; j < i; ++j) 22 C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod; 23 for (register int i = 0; i <= 2335; ++i) f[i][0] = 1; 24 for (register int i = 0; i <= 2335; ++i) 25 for (register int j = 1; j <= 2335; ++j) 26 f[i][j] = (C[i][j] + f[i][j - 1]) % mod; 27 int T; 28 scanf("%d", &T); 29 while (T--) { 30 ll n, k; 31 scanf("%lld%lld", &n, &k); 32 printf("%lld\n", calc(n, k)); 33 } 34 return 0; 35 }
