題意

• 求\(\displaystyle \sum_{i=0}^k {n \choose i} \mod 2333\)

直接推式子：

令\(p=2333,k=tp+r\)

\[\displaystyle \sum_{i=0}^k {n \choose i} \\ = \sum_{i=0}^{tp+r} {n \choose i} \\ =\sum_{i=0}^{tp-1} {n \choose i} + \sum_{i=0}^{r} {n \choose {tp+i}} \\ =\sum_{i=0}^{tp-1} {\frac n p \choose \frac i p} {n \% p \choose i \% p} + \sum_{i=0}^{r} {\frac n p \choose t} {n \% p \choose i} \\ \]

\[=\sum_{i=0}^{tp-1} {\frac n p \choose \frac i p} {n \% p \choose i \% p} + {\frac n p \choose t}\sum_{i=0}^{r} {n \% p \choose i} \\ = \sum_{i=0}^{p-1} {n \% p \choose i} \sum_{j=0}^{t-1} {\frac n p \choose j} + {\frac n p \choose t} \sum_{i=0}^{r} {n \% p \choose i} \\ = 2^{n \% p} \sum_{i=0}^{t-1} {\frac n p \choose i} + {\frac n p \choose t} \sum_{i=0}^{r} {n \% p \choose i} \\ \]

那麼令\(\displaystyle S(n,k) = \sum_{i=0}^k {n \choose i}\),就可以得到：

\[S(n,k)=2^{n \% p} \cdot S(\frac n p, \frac k p -1) + {\frac n p \choose i} \cdot S(n \% p,k \% r) \]

遞迴求解即可

#include<bits/stdc++.h>
#define For(i, a, b) for(int i = (a), en = (b); i <= en; ++i)
#define Rof(i, a, b) for(int i = (a), en = (b); i >= en; --i)
#define Tra(u, i) for(int i = hd[u]; ~i; i = e[i].net)
#define cst const
#define LL long long
#define DD double
#define LD long double
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define inf 0x3f3f3f3f
#define Inf 0x3f3f3f3f3f3f3f3f
#define eps 1e-12
#define mod 2333
using namespace std;

int t, fac[mod + 5], ifac[mod + 5];
LL n, k;

template <class T>
void read(T &x){
	char ch;
	bool ok;
	for(ok = 0, ch = getchar(); !isdigit(ch); ch = getchar()) if(ch == '-') ok = 1;
	for(x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar());
	if(ok) x = -x;
}

int fp(int x, int y){
	int asi = 1;
	while(y){
		if(y & 1) asi = 1ll * asi * x % mod;
		x = 1ll * x * x % mod;
		y >>= 1;
	}
	return asi;
}

int c(int x, int y){return x < y ? 0 : 1ll * fac[x] * ifac[y] % mod * ifac[x - y] % mod;}
int luc(LL x, LL y){
	if(!y) return 1;
	if(!x) return 0;
	return 1ll * c(x % mod, y % mod) * luc(x / mod, y / mod) % mod;
}

int f[mod + 5][mod + 5];
map<LL, map<LL, int> > ma;
int get_s(LL x, LL y){
	if(y < 0) return 0;
	if(x < mod && y < mod) return f[x][y];
	if(ma[x][y]) return ma[x][y] - 1;
	int asi = 1ll * luc(x / mod, y / mod) * get_s(x % mod, y % mod) % mod;
	asi = (asi + 1ll * fp(2, x % mod) * get_s(x / mod, y / mod - 1) % mod) % mod;
	ma[x][y] = asi + 1;
	return asi;
}

int main(){
	//freopen("in", "r", stdin);
	//freopen("c.out", "w", stdout);
	fac[0] = fac[1] = ifac[0] = ifac[1] = 1;
	For(i, 2, mod - 1) fac[i] = 1ll * fac[i - 1] * i % mod;
	ifac[mod - 1] = fp(fac[mod - 1], mod - 2);
	Rof(i, mod - 2, 2) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
	For(i, 0, mod) For(j, 0, mod)
		f[i][j] = (1ll * f[i][j] + (j ? f[i][j - 1] : 0) + c(i, j)) % mod;
	read(t);
	while(t--){
		read(n); read(k);
		printf("%d\n", get_s(n, k));
	}
	return 0;
}