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    @ 2025-8-24 22:08:51

    自动搬运

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    来自洛谷,原作者为

    avatar NaCly_Fish
    北海虽赊,扶摇可接。

    搬运于2025-08-24 22:08:51,当前版本为作者最后更新于2019-03-17 00:20:29,作者可能在搬运后再次修改,您可在原文处查看最新版

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    以下是正文


    大名鼎鼎的欧拉公式

    eix=cosx+isinx\text e^{\text ix}=\cos x+ \text i\sin x

    把这个式子稍微变形一下

    eix=cosxisinx\text e^{-\text ix}=\cos x-\text i\sin x

    将这两个式子相加或相减一下,得到

    2cosx=eix+eix2\cos x= \text e^{\text ix}+ \text e^{-\text ix} 2isinx=eixeix2\text i\sin x=\text e^{ix}-\text e^{-\text ix}

    然后我们移个项,并用 F(x)F(x) 代换 xx,就得出了结果

    $$\cos F(x)=\frac{\text e^{\text iF(x)}+\text e^{-\text iF(x)}}{2} $$$$\sin F(x)=\frac{\text e^{\text iF(x)}-\text e^{-\text iF(x)}}{2\text i} $$

    可是这里还有个虚数单位 i\text i 呢。。这怎么取模啊 QAQ
    别慌,我们冷静分析一下

    虚数单位 i\text i 实际上就是 ω4\omega_4,也就是 g(p1)/4g^{(p-1)/4},在模 pp 下就等于 8658371886583718

    剩下的,就是套上一个多项式指数函数的板子了。
    Code:

    #pragma GCC optimize ("unroll-loops")
    #pragma GCC optimize (2)
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #define N 262147
    #define ll long long
    #define reg register
    #define p 998244353
    #define add(x,y) (x+y>=p?x+y-p:x+y)
    #define dec(x,y) (x<y?x-y+p:x-y)
    using namespace std;
    
    inline int power(int a,int t){
    	int res = 1;
    	while(t){
    		if(t&1) res = (ll)res*a%p;
    		a = (ll)a*a%p;
    		t >>= 1; 
    	}
    	return res;
    }
    
    inline void read(int &x){
        x = 0;
        char c = getchar();
        while(c<'0'||c>'9') c = getchar();
        while(c>='0'&&c<='9'){
            x = (x<<3)+(x<<1)+(c^48);
            c = getchar();
        }
    }
    
    void print(int x){
        if(x>9) print(x/10);
        putchar(x%10+'0');
    }
    
    int rev[N],rt[N],inv[N],inv2[N];
    int siz;
    
    void init(int n){
        int w,lim = 1;
        while(lim<=n) lim <<= 1,++siz;
        for(reg int i=1;i!=lim;++i) rev[i] = (rev[i>>1]>>1)|((i&1)<<(siz-1));
        inv2[1] = inv[1] = rt[lim>>1] = 1;
        w = power(114514,(p-1)>>siz); //114514 也是 998244353 的原根
        for(reg int i=(lim>>1)+1;i!=lim;++i) rt[i] = (ll)rt[i-1]*w%p;
        for(reg int i=(lim>>1)-1;i;--i) rt[i] = rt[i<<1];
        for(reg int i=2;i<=n;++i) inv[i] = (ll)(p-p/i)*inv[p%i]%p;
        for(reg int i=1;i<=siz;++i) inv2[1<<i] = p-((p-1)>>i);
    }
    
    inline int getlen(int n){
        return 1<<(32-__builtin_clz(n));
    }
    
    inline void NTT(int *f,int type,int lim){
        if(type==-1) reverse(f+1,f+lim);
        static unsigned long long a[N];
        reg int x,shift = siz-__builtin_ctz(lim);
        for(reg int i=0;i!=lim;++i) a[rev[i]>>shift] = f[i];
        for(reg int mid=1;mid!=lim;mid<<=1)
        for(reg int j=0;j!=lim;j+=(mid<<1))
        for(reg int k=0;k!=mid;++k){
            x = a[j|k|mid]*rt[mid|k]%p;
            a[j|k|mid] = a[j|k]-x+p;
            a[j|k] += x;
        } 
        for(reg int i=0;i!=lim;++i) f[i] = a[i]%p;
        if(type==1) return;
        x = inv2[lim];
        for(reg int i=0;i!=lim;++i) f[i] = (ll)f[i]*x%p;
    }
    
    inline void inverse(const int *f,int n,int *R){
        static int g[N],h[N],s[30];
        memset(g,0,getlen(n<<1)<<2);
        int lim = 1,top = 0;
        while(n){
            s[++top] = n;
            n >>= 1;
        }
        g[0] = power(f[0],p-2);
        while(top--){
            n = s[top+1];
            while(lim<=(n<<1)) lim <<= 1;
            memcpy(h,f,(n+1)<<2);
            memset(h+n+1,0,(lim-n)<<2);
            NTT(g,1,lim),NTT(h,1,lim);
            for(reg int i=0;i!=lim;++i) g[i] = g[i]*(2-(ll)g[i]*h[i]%p+p)%p;
            NTT(g,-1,lim);
            memset(g+n+1,0,(lim-n)<<2);
        }
        memcpy(R,g,(n+1)<<2);
    }
    
    inline void log(int *f,int n){
        static int g[N];
        int lim = getlen(n<<1);
        inverse(f,n,g);
        memset(g+n+1,0,(lim-n)<<2);
        for(reg int i=0;i!=n;++i) f[i] = (ll)f[i+1]*(i+1)%p;
        f[n] = 0;
        NTT(f,1,lim),NTT(g,1,lim);
        for(reg int i=0;i!=lim;++i) f[i] = (ll)f[i]*g[i]%p;
        NTT(f,-1,lim);
        for(reg int i=n;i;--i) f[i] = (ll)f[i-1]*inv[i]%p;
        f[0] = 0;
        memset(f+n+1,0,(lim-n)<<2);
    }
    
    inline void exp(const int *f,int n,int *R){
        static int g[N],h[N],s[30];
        int lim = 1,top = 0;
        memset(g,0,getlen(n<<1)<<2);
        while(n){
            s[++top] = n;
            n >>= 1;
        }
        g[0] = 1;
        while(top--){
            n = s[top+1];
            while(lim<=(n<<1)) lim <<= 1;
            memcpy(h,g,(n+1)<<2);
            memset(h+n+1,0,(lim-n)<<2);
            log(g,n);
            for(reg int i=0;i<=n;++i) g[i] = dec(f[i],g[i]);
            g[0] = add(g[0],1);
            NTT(g,1,lim),NTT(h,1,lim);
            for(reg int i=0;i!=lim;++i) g[i] = (ll)g[i]*h[i]%p;
            NTT(g,-1,lim);
            memset(g+n+1,0,(lim-n)<<2);
        }
        memcpy(R,g,(n+1)<<2);
    }
    
    #define img 86583718
    
    inline void cos(const int *f,int n,int *R){
        static int g[N],h[N];
        for(reg int i=0;i<=n;++i) g[i] = (ll)f[i]*img%p;
        exp(g,n,g);
        inverse(g,n,h);
        for(reg int i=0;i<=n;++i) R[i] = 499122177ll*(g[i]+h[i])%p;
        
    }
    
    inline void sin(const int *f,int n,int *R){
        static int g[N],h[N];
        for(reg int i=0;i<=n;++i) g[i] = (ll)f[i]*img%p;
        exp(g,n,g);
        inverse(g,n,h);
        int x = power(img<<1,p-2);
        for(reg int i=0;i<=n;++i) R[i] = (ll)x*(g[i]-h[i]+p)%p;
    }
    
    #undef img;
    
    int F[N];
    int n,tp;
    
    int main(){
        read(n),read(tp);
        init(n<<1|1);
        for(reg int i=0;i!=n;++i) read(F[i]);
        if(tp==0) sin(F,n-1,F);
        else cos(F,n-1,F);
        for(reg int i=0;i!=n;++i) print(F[i]),putchar(' ');
    	return 0;	
    }
    
    • 1

    信息

    ID
    4243
    时间
    1000ms
    内存
    125MiB
    难度
    6
    标签
    递交数
    0
    已通过
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