题意
- 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} \
]
那么令(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)
]
递归求解即可
#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
", get_s(n, k));
}
return 0;
}