FFT模板——copied from hzwer

 /*
 Welcome Hacking
 Wish You High Rating
 */
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<ctime>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<string>
 #include<complex>
 using namespace std;
 int read(){
     int xx=,ff=;char ch=getchar();
     while(ch>''||ch<''){if(ch=='-')ff=-;ch=getchar();}
     while(ch>=''&&ch<=''){xx=(xx<<)+(xx<<)+ch-'';ch=getchar();}
     return xx*ff;
 }
 const int maxn=(<<)+;
 const double PI=acos(-1.0);
 typedef complex<double> C;
 int N,M,L,R[maxn];
 C a[maxn],b[maxn];
 void FFT(C a[],int arg){
     for(int i=;i<N;i++)
         if(i<R[i])
             swap(a[i],a[R[i]]);
     for(int i=;i<N;i<<=){
         C wn(cos(PI/i),arg*sin(PI/i));
         for(int p=i<<,j=;j<N;j+=p){
             C w(,);
             for(int k=;k<i;k++,w*=wn){
                 C x=a[j+k],y=w*a[j+k+i];
                 a[j+k]=x+y,a[j+k+i]=x-y;
             }
         }
     }
 }
 int main(){
     //freopen("in","r",stdin);
     N=read(),M=read();
     for(int i=;i<=N;i++)
         a[i]=read();
     for(int i=;i<=M;i++)
         b[i]=read();
     M+=N;
     for(N=;N<=M;N<<=)
         L++;
     for(int i=;i<N;i++)
         R[i]=(R[i>>]>>)|((i&)<<(L-));
     FFT(a,);FFT(b,);
     for(int i=;i<=N;i++)
         a[i]*=b[i];
     FFT(a,-);
     for(int i=;i<=M;i++)
         printf("%d ",(int)(a[i].real()/N+0.5));
     return ;
 }

FFT还有蝶形数组那部分不懂，emmmm，先这样吧，抄一个黄学长的模板，体验acmer用模板切题的快感。

hdu1402

A * B Problem Plus

Problem Description

Calculate A * B.

 

Input

Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

 

Output

For each case, output A * B in one line.

 

Sample Input

1
2
1000
2

 

Sample Output

2
2000

 

Author

DOOM III

 

 

直接上模板改一改，注意清除前导0和进位（错了一次，逃）

 /*
 Welcome Hacking
 Wish You High Rating
 */
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<ctime>
 #include<cstdlib>
 #include<algorithm>
 #include<cmath>
 #include<string>
 #include<complex>
 using namespace std;
 int read(){
     int xx=,ff=;char ch=getchar();
     while(ch>''||ch<''){if(ch=='-')ff=-;ch=getchar();}
     while(ch>=''&&ch<=''){xx=(xx<<)+(xx<<)+ch-'';ch=getchar();}
     return xx*ff;
 }
 const int maxn=(<<)+;
 const double PI=acos(-1.0);
 typedef complex<double> C;
 int N,M,L,R[maxn];
 C a[maxn],b[maxn];
 void FFT(C a[],int arg){
     for(int i=;i<N;i++)
         if(i<R[i])
             swap(a[i],a[R[i]]);
     for(int i=;i<N;i<<=){
         C wn(cos(PI/i),arg*sin(PI/i));
         for(int p=i<<,j=;j<N;j+=p){
             C w(,);
             for(int k=;k<i;k++,w*=wn){
                 C x=a[j+k],y=w*a[j+k+i];
                 a[j+k]=x+y,a[j+k+i]=x-y;
             }
         }
     }
 }
 char s1[],s2[];
 int ans[];
 int main(){
     //freopen("in","r",stdin);
     /*N=read(),M=read();
     for(int i=0;i<=N;i++)
         a[i]=read();
     for(int i=0;i<=M;i++)
         b[i]=read();*/
     while(scanf("%s",s1)!=EOF){
         scanf("%s",s2);
         N=strlen(s1)-,M=strlen(s2)-;
         for(int i=;i<=N;i++)
             a[i]=s1[N-i]-'';
         for(int i=;i<=M;i++)
             b[i]=s2[M-i]-'';
         M+=N;
         for(N=;N<=M;N<<=)
             L++;
         for(int i=;i<N;i++)
             R[i]=(R[i>>]>>)|((i&)<<(L-));
         FFT(a,);FFT(b,);
         for(int i=;i<=N;i++)
             a[i]*=b[i];
         FFT(a,-);
         for(int i=M;i>=;i--)
             ans[i]=(int)(a[i].real()/N+0.5);
         int k;
         for(k=;k<=M||ans[k];k++)
             if(ans[k]>=)
                 ans[k+]+=ans[k]/,ans[k]%=;
         while(!ans[k]&&k>)
             k--;
         for(int i=k;i>=;i--)
             printf("%d",ans[i]);
         puts("");
         memset(a,,sizeof(a));
         memset(b,,sizeof(b));
         memset(R,,sizeof(R));
         memset(ans,,sizeof(ans));
         L=;
     }
     return ;
 }