BZOJ 2179FFT快速傅立叶

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2179

题目大意：给出两个n位10进制整数x和y，你需要计算x*y。

题解：FFT，不会的可以膜拜陈老师（非clj）QQ：297086016

代码：

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #define inf 1<<30
 #define maxm 270005
 #define pi acos(-1)
 using namespace std;
 struct F{
     double rea,ima;
     F operator +(const F &x){return (F){rea+x.rea,ima+x.ima};}
     F operator -(const F &x){return (F){rea-x.rea,ima-x.ima};}
     F operator *(const F &x){return (F){rea*x.rea-ima*x.ima,rea*x.ima+ima*x.rea};}
 }a[maxm],b[maxm],c[maxm],w,wn,t1,t2;
 int n,m,l,ans[maxm],rev[maxm];
 void read(F *a)
 {
     char ch;
     while (ch=getchar(),ch<''||ch>'');
     for (int i=m-; ch>=''&&ch<='';ch=getchar(),i--) a[i].rea=ch-'';
 }
 int re(int v)
 {
     int t=;
     for (int i=; i<l; i++) t<<=,t|=v&,v>>=;
     return t;
 }
 void fft(F *a,int type)
 {
     for (int i=; i<n; i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
     for (int s=; s<=n; s<<=)
     {
         wn=(F){cos(type**pi/s),sin(type**pi/s)};
         for (int i=; i<n; i+=s)
         {
             w=(F){,};
             for (int j=i; j<(i+(s>>)); j++,w=w*wn)
             {
                 t1=a[j],t2=w*a[j+(s>>)];
                 a[j]=t1+t2,a[j+(s>>)]=t1-t2;
             }
         }
     }
 }
 int main()
 {
     scanf("%d\n",&m);
     for (n=; n<(m<<); n<<=) l++;
     for (int i=; i<n; i++) rev[i]=re(i);
     read(a); read(b);
     fft(a,); fft(b,);
     for (int i=; i<n; i++) c[i]=a[i]*b[i];
     fft(c,-);
     for(int i=;i<n;i++) ans[i]=int(c[i].rea/n+0.5);
     for(int i=;i<n;i++) ans[i+]+=ans[i]/,ans[i]=ans[i]%;
     int pps=n-;while(ans[pps]==&&pps)pps--;
     for(;pps>=;pps--)printf("%d",ans[pps]);printf("\n");
     return ;
 }