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来自洛谷,原作者为

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自动搬运只会搬运当前题目点赞数最高的题解,您可前往洛谷题解查看更多
以下是正文
大名鼎鼎的欧拉公式
把这个式子稍微变形一下
将这两个式子相加或相减一下,得到
然后我们移个项,并用 代换 ,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 呢。。这怎么取模啊 QAQ
别慌,我们冷静分析一下

虚数单位 实际上就是 ,也就是 ,在模 下就等于 。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
- 0
- 上传者