Dyt's math [数学, 推式子]

$Dyt's math$

$求解下式,对 998244353 取模:$

$sum_{i=1}^n C_i^k b^i$

$k le 5 imes 10^5, n le 10^{18},2 le b le 998244353.$

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$color{red}{正解部分}$

设 $F_k = sumlimits_{i=1}^n C_i^k b^i$, 则 $F_{k-1} = sumlimits_{i=1}^n C_i^{k-1} b^i$,

$bF_{k-1} = sumlimits_{i=2}^{n+1} C_{i-1}^{k-1} b^i$,

$bF_k = sumlimits_{i=2}^{n+1} C_{i-1}^k b^i$,

$herefore bF_{k-1}+bF_k = sumlimits_{i=2}^{n+1}left(C_{i-1}^k +C_{i-1}^{k-1} ight) b^i = sumlimits_{i=2}^{n+1}C_i^k b^i = F_k + C_{n+1}^kb^{n+1}-C_1^kb$ .

$herefore F_k = frac{bF_{k-1}-C_{n+1}^kb^{n+1}+C_1^kb}{1-b}$

$又ecause F_0 = sum_{i=1}^n b^i = frac{b(1-b^n)}{1-b}$

所以可以 $O(K)$ 得到 $F_k$ .

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$color{red}{实现部分}$

#include<bits/stdc++.h>
#define reg register
typedef long long ll;

const int maxn = 500005;
const int mod = 998244353;

int K;
int B;

ll N;

int Ksm(int a, ll b){int s=1;while(b){if(b&1)s=1ll*s*a%mod;a=1ll*a*a%mod;b>>=1;}return s;}

int main(){
        scanf("%lld%d%d", &N, &B, &K);
        int inv_b = Ksm(B-1, mod-2);
        int Ans = 1ll*B*(Ksm(B, N)-1)%mod*inv_b % mod;
        int C = (N + 1)%mod;
        for(reg int i = 1; i <= K; i ++){
                int C1kb = (i==1)*B;
                Ans = -1ll*B*Ans % mod;
                Ans += 1ll*C*Ksm(B, N+1)%mod;
                Ans -= C1kb;
                Ans %= mod, Ans += mod, Ans %= mod;
                Ans = 1ll*Ans*inv_b % mod;
                C = (N-i+1)%mod*C%mod*Ksm(i+1, mod-2)%mod;
        }
        printf("%d
", Ans);
        return 0;
}