P4512 【模板】多项式除法

P4512 【模板】多项式除法

链接

分析　　

　　多项式除法

注意的地方：

75，76行开始时是这样写的：

memcpy(TA,a,sizeof(int)*(n+1));memset(TA+n+1,0,sizeof(TA));
memcpy(TB,b,sizeof(int)*(m+1));memset(TB+m+1,0,sizeof(TB));

然后开O2的情况不过。最后发现时后面的memset不能这样写。然后在本地开O2测试，可以过样例。。。 ~ 惊！~ 吓！

代码

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cctype>

#define P 998244353
#define G 3
#define Gi 332748118 
#define N 270000

using namespace std;

int A[N],B[N],D[N],TA[N],TB[N],DR[N],Binv[N],R[N];

inline int read() {
    int x = 0,f = 1;char ch=getchar();
    for (; !isdigit(ch); ch=getchar()) if(ch=='-')f=-1;
    for (; isdigit(ch); ch=getchar()) x=x*10+ch-'0';
    return x*f;
}

int ksm(int a,int b) {
    int ans = 1;
    while (b) {
        if (b & 1) ans = (1ll * ans * a) % P;
        a = (1ll * a * a) % P;
        b >>= 1;
    }
    return ans;
}
void NTT(int *a,int n,int ty) {
    for (int i=0,j=0; i<n; ++i) {
        if (i < j) swap(a[i],a[j]);
        for (int k=(n>>1); (j^=k)<k; k>>=1);
    }
    for (int w1,m=2; m<=n; m<<=1) {
        if (ty == 1) w1 = ksm(G,(P-1)/m);
        else w1 = ksm(Gi,(P-1)/m);
        for (int i=0; i<n; i+=m) {
            int w = 1;
            for (int k=0; k<(m>>1); ++k) {
                int u = a[i+k],t = 1ll * w * a[i+k+(m>>1)] % P;
                a[i+k] = (u + t) % P;
                a[i+k+(m>>1)] = (u - t + P) % P;
                w = 1ll * w * w1 % P;
            }
        }
    }
    if (ty==-1) {
        int inv = ksm(n,P-2);
        for (int i=0; i<n; ++i) a[i] = 1ll * a[i] * inv % P;
    }
    
}
void Inv(int *A,int *B,int n) { 
    int len = 1;
    while (len <= n) len <<= 1; 
    B[0] = ksm(A[0],P-2);
    for (int m=2; m<=len; m<<=1) {
        for (int i=0; i<m; ++i) TA[i] = A[i],TB[i] = B[i];
        NTT(TA,m<<1,1);
        NTT(TB,m<<1,1);
        for (int i=0; i<(m<<1); ++i) TA[i] = 1ll*TA[i]*TB[i]%P*TB[i]%P;
        NTT(TA,m<<1,-1);
        for (int i=0; i<m; ++i) B[i] = (1ll*2*B[i]%P-TA[i]+P)%P;
    }
    memset(TA,0,sizeof(TA));
    memset(TB,0,sizeof(TB));
}
void Mul(int *a,int *b,int *C,int n,int m) {
    int len = 1;
    while (len <= n+m) len <<= 1;
    for (int i=0; i<=n; ++i) TA[i] = a[i];
    for (int i=0; i<=m; ++i) TB[i] = b[i];
    NTT(TA,len,1);
    NTT(TB,len,1);
    for (int i=0; i<len; ++i) C[i] = (1ll * TA[i] * TB[i]) % P,TA[i] = TB[i] = 0;
    NTT(C,len,-1);
}
int main() {
    int n = read() ,m = read() ;
    for (int i=0; i<=n; ++i) A[i] = read();
    for (int i=0; i<=m; ++i) B[i] = read();
    
    reverse(A,A+n+1);reverse(B,B+m+1); 
    
    Inv(B,Binv,n-m); // 求B转换后的逆元 
    Mul(A,Binv,D,n-m,n-m); // 求转换后的D 
    reverse(D,D+n-m+1);  
    for (int i=0; i<=n-m; ++i) printf("%d ",D[i]);puts("");
    
    reverse(A,A+n+1);reverse(B,B+m+1);
    Mul(D,B,R,n-m,m); // 求D*B 
    for (int i=0; i<m; ++i) R[i] = (A[i] - R[i] + P) % P; // 求R 
    for (int i=0; i<m; ++i) printf("%d ",R[i]);
    
    return 0;
}