求助 倍增快速幂36pts WA了一堆
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求助 倍增快速幂36pts WA了一堆
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SenriAkane楼主2022/8/10 11:08
//#pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N = 1e6+7;
const int mod = 998244353;

template <typename T>void read(T &x)
{
    x = 0;
    register int f = 1;
    register char ch = getchar();
    while(ch < '0' || ch > '9') {if(ch == '-')f = -1;ch = getchar();}
    while(ch >= '0' && ch <= '9') {x = x * 10 + ch - '0';ch = getchar();}
    x *= f;
}

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

namespace Poly
{
    #define mul(x, y) (1ll * x * y >= mod ? 1ll * x * y % mod : 1ll * x * y)
    #define minus(x, y) (1ll * x - y < 0 ? 1ll * x - y + mod : 1ll * x - y)
    #define plus(x, y) (1ll * x + y >= mod ? 1ll * x + y - mod : 1ll * x + y)
    #define ck(x) (x >= mod ? x - mod : x)//取模运算太慢了

    typedef vector<int> poly;
    const int G = 3;//根据具体的模数而定,原根可不一定不一样!!!
    //一般模数的原根为 2 3 5 7 10 6
    const int inv_G = qpow(G, mod - 2);
    int RR[N], deer[2][20][N], inv[N];

    void init(const int t) {//预处理出来NTT里需要的w和wn,砍掉了一个log的时间
        for(int p = 1; p <= t; ++ p) {
            int buf1 = qpow(G, (mod - 1) / (1 << p));
            int buf0 = qpow(inv_G, (mod - 1) / (1 << p));
            deer[0][p][0] = deer[1][p][0] = 1;
            for(int i = 1; i < (1 << p); ++ i) {
                deer[0][p][i] = 1ll * deer[0][p][i - 1] * buf0 % mod;//逆
                deer[1][p][i] = 1ll * deer[1][p][i - 1] * buf1 % mod;
            }
        }
        inv[1] = 1;
        for(int i = 2; i <= (1 << t); ++ i)
            inv[i] = 1ll * inv[mod % i] * (mod - mod / i) % mod;
    }

    int NTT_init(int n) {//快速数论变换预处理
        int limit = 1, L = 0;
        while(limit <= n) limit <<= 1, L ++ ;
        for(int i = 0; i < limit; ++ i)
            RR[i] = (RR[i >> 1] >> 1) | ((i & 1) << (L - 1));
        return limit;
    }

    void NTT(poly &A, int type, int limit) {//快速数论变换
        A.resize(limit);
        for(int i = 0; i < limit; ++ i)
            if(i < RR[i])
                swap(A[i], A[RR[i]]);
        for(int mid = 2, j = 1; mid <= limit; mid <<= 1, ++ j) {
            int len = mid >> 1;
            for(int pos = 0; pos < limit; pos += mid) {
                int *wn = deer[type][j];
                for(int i = pos; i < pos + len; ++ i, ++ wn) {
                    int tmp = 1ll * (*wn) * A[i + len] % mod;
                    A[i + len] = ck(A[i] - tmp + mod);
                    A[i] = ck(A[i] + tmp);
                }
            }
        }
        if(type == 0) {
            for(int i = 0; i < limit; ++ i)
                A[i] = 1ll * A[i] * inv[limit] % mod;
        }
    }

    poly poly_mul(poly A, poly B) {//多项式乘法
        int deg = A.size() + B.size() - 1;
        int limit = NTT_init(deg);
        poly C(limit);
        NTT(A, 1, limit);
        NTT(B, 1, limit);
        for(int i = 0; i < limit; ++ i)
            C[i] = 1ll * A[i] * B[i] % mod;
        NTT(C, 0, limit);
        C.resize(deg);
        return C;
    }
    void update(poly &f,int deg) {
    	int len = f.size();
    	f.resize(deg);
    	if (len<deg) {
			for (int i=len;i<deg;i++) f[i] = 0;
    	}
    }
    poly poly_pow(poly f, int k,int deg) {//多项式快速幂
 		poly res(deg);
 		for (int i=0;i<deg;i++) res[i] = 0;
 		res[0] = 1;
 		while (k) {
 			if (k&1) {
 				res = poly_mul(res,f);
 				update(res,deg);
 			}
 			f = poly_mul(f,f);
 			update(f,deg);
 			k >>= 1;
 		}
 		update(res,deg);
 		return res;
    }
	
}
using namespace Poly;
int main()
{
    Poly::init(19);
	int n;
	string s;
	int k = 0;
	cin>>n>>s;
	for (int i=0;i<s.length();i++) {
		k = (1ll*k*10%mod+s[i]-'0')%mod;
	}
	poly f(n+5);
	for (int i=0;i<n;i++) {
		cin>>f[i];
	}
	f = poly_pow(f,k,n+5);
	for (int i=0;i<n;i++) cout<<f[i]<<" ";
}
2022/8/10 11:08
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