[洛谷P4345][SHOI2015]超能粒子炮·改

题目大意：给你$n,k$，求：
$$
\sum\limits_{i=0}^k\binom n i\pmod{2333}
$$
题解：令$p=2333,f(n,k)\equiv\sum\limits_{i=0}^k\binom n i\pmod p$
$$
\begin{align*}
f(n,k)\equiv&\sum\limits_{i=0}^k\binom n i\pmod p\\    \equiv&\sum\limits_{i=0}^k\binom{\big\lfloor\frac np\big\rfloor}{\big\lfloor\frac ip\big\rfloor}\binom{n\bmod p}{i\bmod p}\pmod p\\
\end{align*}\\
令s=\left\lfloor\dfrac k p\right\rfloor
$$

$$
\begin{align*}
f(n,k)\equiv&[\sum\limits_{i=0}^{p-1}\binom{n\bmod p}{i}][\sum\limits_{i=0}^{s-1}\binom{\big\lfloor\frac n p\big\rfloor}{i}]\\
    &+\binom{\left\lfloor\frac np\right\rfloor}{s}\sum\limits_{i=sp}^k\binom{n\bmod p}{i\bmod p}\pmod p\\
    \equiv&[\sum\limits_{i=0}^{p-1}\binom{n\bmod p}{i}][\sum\limits_{i=0}^{s-1}\binom{\big\lfloor\frac n p\big\rfloor}{i}]\\
    &+\binom{\left\lfloor\frac np\right\rfloor}{s}\sum\limits_{i=0}^{k\bmod p}\binom{n\bmod p}{i}\pmod p\\
    \equiv&f(n\bmod p, p-1)f(\left\lfloor\dfrac np\right\rfloor,s-1)\\
    &+\binom{\big\lfloor\frac np\big\rfloor}{s}f(n\bmod p,k\bmod p)\pmod p\\
\end{align*}
$$

卡点：未注意$n,k\leqslant10^{18}$



C++ Code：

#include <cstdio>
const int mod = 2333;
#define maxn mod
inline void reduce(int &x) { x += x >> 31 & mod; }

int Tim;
long long n, k;
int com[maxn][maxn], pre[maxn][maxn];

int C(long long a, long long b) {
	if (a < b) return 0;
	if (a < mod) return com[a][b];
	return com[a % mod][b % mod] * C(a / mod, b / mod) % mod;
}
int solve(long long n, long long k) {
	if (k < 0) return 0;
	if (n < mod && k < mod) return pre[n][k];
	const long long s = k / mod;
	return (pre[n % mod][mod - 1] * solve(n / mod, s - 1) + pre[n % mod][k % mod] * C(n / mod, s)) % mod;
}
int main() {
	scanf("%d", &Tim);
	for (int i = 0; i < mod; ++i) {
		*com[i] = *pre[i] = 1;
		for (int j = 1; j <= i; ++j) {
			reduce(com[i][j] = com[i - 1][j] + com[i - 1][j - 1] - mod);
			reduce(pre[i][j] = pre[i][j - 1] + com[i][j] - mod);
		}
		for (int j = i + 1; j < mod; ++j) reduce(pre[i][j] = pre[i][j - 1] + com[i][j] - mod);
	}
	while (Tim --> 0) {
		scanf("%lld%lld", &n, &k);
		printf("%d\n", solve(n, k));
	}
	return 0;
}