既然这题模数是今天日期减去(7 imes 10^5)，那就要赶紧把这题做了。
首先肯定是考虑指数型生成函数，列出来之后使用单位根反演一波。
[egin{aligned}Ans&=(sum_{d|i}frac{x^i}{i!})^k \&=(frac{sum_{j=0}^{d-1}sum_{i=0}^nfrac{(omega_d^jx)^i}{i!}}{d})^k \&=(frac{sum_{i=0}^{d-1}e^{omega_d^ix}}{d})^k \&=frac{1}{d^k}(sum_{i=0}^{d-1}e^{omega_d^ix})^kend{aligned}
]
当(d=1)时
[Ans=e^{kx}
]
当(d=2)时
[egin{aligned}Ans&=frac{1}{2^k}(e^x+e^{-x})^k \&=frac{1}{2^k}sum_{i=0}^kinom{k}{i}e^{(2i-k)x}end{aligned}
]
当(d=3)时
[egin{aligned}Ans&=frac{1}{3^k}(e^x+e^{omega x}+e^{aromega x})^k \&=frac{1}{3^k}sum_{i=0}^ksum_{j=0}^{k-i}inom{k}{i,j}e^{(i+jomega+(k-i-j)aromega)x}end{aligned}
]
时间复杂度(O(k^{d-1}log n))
#include<bits/stdc++.h>
#define Rint register int
using namespace std;
typedef long long LL;
const int N = 500003, mod = 19491001, w = 18827933, w2 = (LL) w * w % mod;
int n, k, d, fac[N], inv[N], ans;
inline void upd(int &a, int b){a += b; if(a >= mod) a -= mod;}
inline int kasumi(int a, int b){
int res = 1;
while(b){
if(b & 1) res = (LL) res * a % mod;
a = (LL) a * a % mod; b >>= 1;
}
return res;
}
int main(){
scanf("%d%d%d", &n, &k, &d); n %= mod - 1;
if(d == 1){printf("%d", kasumi(k, n)); return 0;}
fac[0] = 1;
for(Rint i = 1;i <= k;i ++) fac[i] = (LL) fac[i - 1] * i % mod;
inv[k] = kasumi(fac[k], mod - 2);
for(Rint i = k;i;i --) inv[i - 1] = (LL) inv[i] * i % mod;
if(d == 2){
for(Rint i = 0;i <= k;i ++)
upd(ans, (LL) inv[i] * inv[k - i] % mod * kasumi((2 * i - k + mod) % mod, n) % mod);
printf("%d", (LL) ans * kasumi(2, mod - 1 - k) % mod * fac[k] % mod);
} else if(d == 3){
for(Rint i = 0;i <= k;i ++)
for(Rint j = 0;i + j <= k;j ++)
upd(ans, (LL) inv[i] % mod * inv[j] % mod * inv[k - i - j] % mod *
kasumi((i + (LL) j * w % mod + (LL) (k - i - j) * w2 % mod) % mod, n) % mod);
printf("%d", (LL) ans * kasumi(3, mod - 1 - k) % mod * fac[k] % mod);
}
}
EI对于这题还有一个加强版，但是我不会做，先咕了。