(Good morning, notes in a comment.)
Polynomial convolution template:
FFT：
#include<iostream> #include<cstdio> #include<cmath> using namespace std; const int N=3e6+10; double pi=acos(-1); int n,m; struct node{ double x,y; node(double a=0,double b=0){ x = a, y = b; } node operator + (node const &u) const{ return node(x+u.x,y+u.y); } node operator - (node const &u) const{ return node(x-u.x,y-u.y); } node operator * (node const &u) const{ return node(x*u.x-y*u.y,y*u.x+x*u.y); } }f[N]; int pos[N]; void fft(node *f,bool flag){ for(int i=0;i<n;i++){ if(i<pos[i])swap(f[i],f[pos[i]]); } for(int p=2;p<=n;p<<=1){ int len=p>>1; node for (cos ( 2 * pi / p), I ( 2 * pi / p)); if (! flag) fir.y * = - 1 ; for ( int k = 0 ; k <n; k + = p) { node buf(1,0); for(int l=k;l<k+len;l++){ node tt = f * buf [len + l]; f [len + l] = f [the] - tt; in [] = f [t] + tt; buf=buf*fir; } } } } int main () { scanf("%d%d",&n,&m); for(int i=0;i<=n;i++)scanf("%lf",&f[i].x); for(int i=0;i<=m;i++)scanf("%lf",&f[i].y); for(m+=n,n=1;n<=m;n<<=1); for(int i=0;i<n;i++){ pos[i]=(pos[i>>1]>>1)|((i&1)?n>>1:0); } fft(f,1); for(int i=0;i<n;i++)f[i]=f[i]*f[i]; fft(f,0); for(int i=0;i<=m;i++)printf("%d ",(int)(f[i].y/n/2+0.5)); return 0; } // the FFT compare throw accuracy, if need convolution polynomial coefficients range that much difference, the card will be precision // three times greater precision optimization becomes twice the upper limit of the span involved, serious off accuracy
NTT:
#include<iostream> #include<cstdio> using namespace std; const int N=3e6+10,mod=998244353,G=3; int n,m,pos[N]; long long f[N],g[N],invn,invG; long long pw(long long x,long long k){ long long num=1; while(k){ if(k&1)num=num*x%mod; x=x*x%mod; k>>=1; } return num; } void ntt(long long *f,bool flag){ for(int i=0;i<n;i++){ if(i<pos[i])swap(f[i],f[pos[i]]); } for(int p=2;p<=n;p<<=1){ int len=p>>1; long long fir=pw((flag?G:invG),(mod-1)/p); for(int i=0;i<n;i+=p){ long long bur=1; for(int l=i;l<i+len;l++){ long long tt=bur*f[l+len]%mod; f [l + len] = ((f [l] -tT)% v + v)% v; f[l]=(f[l]+tt)%mod; bur = bur *% for mod; } } } } int main () { scanf("%d%d",&n,&m); for(int i=0;i<=n;i++)scanf("%lld",&f[i]); for(int i=0;i<=m;i++)scanf("%lld",&g[i]); for(m+=n,n=1;n<=m;n<<=1); for(int i=0;i<n;i++){ pos[i]=(pos[i>>1]>>1)|((i&1)?(n>>1):0); } invn=pw(n,mod-2),invG=pw(G,mod-2); ntt(f,1),ntt(g,1); for(int i=0;i<n;i++){ f[i]=f[i]*g[i]%mod; } ntt(f,0); for(int i=0;i<=m;i++){ printf("%lld ",f[i]*invn%mod); } return 0; } // array needs to open more than a power of 2, not twice // first unit 1 starts from the root // MOD magnitude of the largest coefficient in the above open out to prevent the mold
Inverse polynomial:
#include<iostream> #include<cstdio> #define ll long long using namespace std; const int N=3e5+10,mod=998244353,G=3; int n,pos[N]; ll a[N],b[N],c[N],invG; ll pw(ll x,ll k){ ll num=1; while(k){ if(k&1)num=num*x%mod; x=x*x%mod; k>>=1; } return num; } void ntt(ll *f,int n,bool flag){ for(int i=0;i<n;i++)if(i<pos[i])swap(f[i],f[pos[i]]); for(int p=2;p<=n;p<<=1){ int len=p>>1; ll fir=pw((flag?G:invG),(mod-1)/p); for(int k=0;k<n;k+=p){ ll bur=1; for(int l=k;l<len+k;l++){ ll tt = f [l + len] * cage% mod; f [len + l] = ((f [l] -tT)% v + v)% v; f[l]=(f[l]+tt)%mod; bur = bur *% for mod; } } } } void getinv(int now,ll *a,ll*b){ if(now==1){b[0]=pw(a[0],mod-2);return;} getinv((now+1)>>1,a,b); int goal=1; while(goal<(now<<1))goal<<=1; ll invn=pw(goal,mod-2); for(int i=0;i<goal;i++)pos[i]=(pos[i>>1]>>1)|((i&1)?(goal>>1):0); for(int i=0;i<now;i++)c[i]=a[i]; for(int i=now;i<goal;i++)c[i]=0; ntt(c,goal,1),ntt(b,goal,1); for(int i=0;i<goal;i++)b[i]=((2ll-c[i]*b[i]%mod)%mod+mod)%mod*b[i]%mod; ntt(b,goal,0); for(int i=0;i<now;i++)b[i]=b[i]*invn%mod; for(int i=now;i<goal;i++)b[i]=0; } int main () { scanf("%d",&n); for(int i=0;i<n;i++)scanf("%lld",&a[i]); invG=pw(G,mod-2); getinv(n,a,b); for(int i=0;i<n;i++)printf("%lld ",b[i]); return 0; } // recursive nature of each layer x ^ now modulo inverse redundant array b After each treatment should be partially emptied out of the mold.
Some of the topics:
Sustained finished up, please remember your own review.