【BZOJ5300】[CQOI2018]九连环 （高精度，FFT）

【BZOJ5300】[CQOI2018]九连环 （高精度，FFT）

题面

BZOJ
洛谷

题解

去这里看吧，多么好

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
#define MAX 150000
const double Pi=acos(-1);
inline int read()
{
    int x=0;bool t=false;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    if(ch=='-')t=true,ch=getchar();
    while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
    return t?-x:x;
}
struct Complex{double a,b;}A[MAX],B[MAX],W[MAX];
Complex operator+(Complex a,Complex b){return (Complex){a.a+b.a,a.b+b.b};}
Complex operator-(Complex a,Complex b){return (Complex){a.a-b.a,a.b-b.b};}
Complex operator*(Complex a,Complex b){return (Complex){a.a*b.a-a.b*b.b,a.b*b.a+a.a*b.b};}
int r[MAX],N,l=0;
void FFT(Complex *P,int N,int opt)
{
    for(int i=0;i<N;++i)if(i<r[i])swap(P[i],P[r[i]]);
    for(int i=1;i<N;i<<=1)
        for(int p=i<<1,j=0;j<N;j+=p)
            for(int k=0;k<i;++k)
            {
                Complex w=W[N/i*k];w.b*=opt;
                Complex X=P[j+k],Y=P[i+j+k]*w;
                P[j+k]=X+Y;P[i+j+k]=X-Y;
            }
    if(opt==-1)for(int i=0;i<N;++i)P[i].a/=N;
}
int Multi(int *a,int *b,int n,int m,int *c)
{
    l=0;for(N=1;N<=n+m;N<<=1)++l;
    for(int i=0;i<N;++i)r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
    for(int i=1;i<N;i<<=1)
        for(int k=0;k<i;++k)
            W[N/i*k]=(Complex){cos(Pi/i*k),sin(Pi/i*k)};
    for(int i=0;i<=n;++i)A[i].a=a[i];
    for(int i=0;i<=m;++i)B[i].a=b[i];
    FFT(A,N,1);FFT(B,N,1);
    for(int i=0;i<N;++i)A[i]=A[i]*B[i];
    FFT(A,N,-1);
    for(int i=0;i<=n+m;++i)c[i]=A[i].a+0.5;
    for(int i=0;i<N;++i)A[i]=B[i]=(Complex){0,0};
    int len=n+m;
    for(int i=0;i<=len;++i)c[i+1]+=c[i]/10,c[i]%=10;
    while(c[len+1])++len,c[len+1]+=c[len]/10,c[len]%=10;
    return len;
}
int a[MAX],b[MAX],c[MAX];
int main()
{
    int T=read();
    while(T--)
    {
        int n=read()+1;
        memset(a,0,sizeof(a));memset(b,0,sizeof(b));
        int la=0,lb=0;a[0]=1;b[0]=2;
        while(n)
        {
            if(n&1)la=Multi(a,b,la,lb,a);
            lb=Multi(b,b,lb,lb,b);
            n>>=1;
        }
        bool o=false;int s=0;
        for(int i=la;~i;--i)
        {
            s=s*10+a[i];
            if(s<3&&!o)continue;
            o=true;printf("%d",s/3);s%=3;
        }
        puts("");
    }
}