Loj#6485. LJJ 學二項式定理(機關根反演)
題目描述
題目描述
題意:求下面式子的答案QAQ。
[ ∑ ( ( n i ) ⋅ s i ⋅ a i m o d 4 ) ] m o d 998244353 [\sum(\tbinom{n}{i}\cdot s^i \cdot a_{i\;\;mod\;\;4}) ]\;mod\;\;998244353 [∑((in)⋅si⋅aimod4)]mod998244353
Solution
The first 機關根反演題 of me。
題目中的式子很有趣,它的形式為 ∑ a i m o d 4 \sum a_{i \;\;mod \;\;4} ∑aimod4
這就是一個典型的機關根反演的形式,是以考慮機關根反演的公式
1 n ∑ i = 0 n − 1 ω i k = [ n ∣ k ] \frac{1}{n}\sum_{i=0}^{n-1} \omega^{ik}=[n|k] n1i=0∑n−1ωik=[n∣k]
我們枚舉 k = i % 4 k=i\%4 k=i%4,原式變為:
∑ k = 0 3 a k ∑ i = 0 n [ 4 ∣ i + 4 − k ] s i ⋅ ( n i ) \sum_{k=0}^3 a_k \sum_{i=0}^n [4|i+4-k]s^i \cdot \tbinom{n}{i} k=0∑3aki=0∑n[4∣i+4−k]si⋅(in)
機關根反演,替換 [ 4 ∣ i + 4 − k ] [4|i+4-k] [4∣i+4−k]進一步化簡得到
∑ k = 0 3 a k ∑ i = 0 n ∑ j = 0 3 ω 4 j ( i + 4 − k ) s i ⋅ ( n i ) \sum_{k=0}^3 a_k \sum_{i=0}^n \sum_{j=0}^3 \omega_4^{j(i+4-k)}s^i \cdot \tbinom{n}{i} k=0∑3aki=0∑nj=0∑3ω4j(i+4−k)si⋅(in)
整理式子:
∑ k = 0 3 a k ∑ j = 0 3 ω 4 j ( 4 − k ) ( ∑ i = 0 n ω 4 i j s i ⋅ ( n i ) ) \sum_{k=0}^3 a_k \sum_{j=0}^3 \omega_4^{j(4-k)} (\sum_{i=0}^n\omega_4^{ij}s^i \cdot \tbinom{n}{i}) k=0∑3akj=0∑3ω4j(4−k)(i=0∑nω4ijsi⋅(in))
二項式定理一波走:
∑ k = 0 3 a k ∑ j = 0 3 ω 4 j ( 4 − k ) ( s ω j + 1 ) n \sum_{k=0}^3 a_k \sum_{j=0}^3 \omega_4^{j(4-k)}(s\omega^j+1)^n k=0∑3akj=0∑3ω4j(4−k)(sωj+1)n
是以我們隻需要預處理之後計算即可。
單次複雜度 O ( c + l g n ) O(c+lgn) O(c+lgn), c c c為常數。
注意 l o n g l o n g long\;\;long longlong,下面的代碼 d e f i n e i n t l l define\;\;int\;\;ll defineintll了。
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>
#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se second
#define int ll
using namespace std;
template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }
typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;
const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{
int f=1,x=0; char c=getchar();
while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }
while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }
return x*f;
}
inline int quick_pow(int x,int y)
{
if (y==0) return 1;
int q=quick_pow(x,y>>1);
return (y&1)?1ll*q*q%mods*x%mods:1ll*q*q%mods;
}
int a[4],w[4],P[4];
signed main() {
int Case=read();
int wn=quick_pow(3,(mods-1)/4);
int inv4=quick_pow(4,mods-2);
w[0]=1;
for (int i=1;i<4;i++) w[i]=1ll*w[i-1]*wn%mods;
while (Case--) {
ll n=read(),s=read(),ans=0;
for (int i=0;i<4;i++) a[i]=read();
for (int i=0;i<4;i++) P[i]=quick_pow((1ll*w[i]*s+1)%mods,n);
for (int i=0;i<4;i++) {
int p=1ll*a[i]*inv4%mods;
for (int j=0;j<4;j++)
ans=(ans+1ll*p*w[j*(4-i)%4]%mods*P[j]%mods)%mods;
}
printf("%d\n",ans);
}
return 0;
}