题意：
思路：FFT模板，自带10倍常数
1 type cp=record 2 x,y:double; 3 end; 4 arr=array[0..510000]of cp; 5 var a,b,cur:arr; 6 n,m,n1,n2,i,j:longint; 7 x:double; 8 9 function jia(a,b:cp;f:longint):cp; 10 begin 11 if f=-1 then 12 begin 13 b.x:=-b.x; b.y:=-b.y; 14 end; 15 jia.x:=a.x+b.x; 16 jia.y:=a.y+b.y; 17 end; 18 19 function mult(a,b:cp):cp; 20 begin 21 mult.x:=a.x*b.x-a.y*b.y; 22 mult.y:=a.x*b.y+a.y*b.x; 23 end; 24 25 procedure swap(var x,y:cp); 26 var t:cp; 27 begin 28 t:=x; x:=y; y:=t; 29 end; 30 31 function max(x,y:longint):longint; 32 begin 33 if x>y then exit(x); 34 exit(y); 35 end; 36 37 procedure fft(var a:arr;n,f:longint); 38 var i,j,k,m:longint; 39 w,u,v:cp; 40 41 begin 42 i:=n>>1; j:=1; 43 while j<n do 44 begin 45 if i<j then swap(a[i],a[j]); 46 k:=n>>1; 47 while k and i>0 do 48 begin 49 i:=i xor k; 50 k:=k>>1; 51 end; 52 i:=i xor k; 53 inc(j); 54 end; 55 m:=2; 56 while m<=n do 57 begin 58 w.x:=cos(2*pi*f/m); w.y:=sin(2*pi*f/m); 59 cur[0].x:=1; cur[0].y:=0; 60 for i:=1 to m-1 do cur[i]:=mult(cur[i-1],w); 61 i:=0; 62 while i<n do 63 begin 64 j:=i; 65 while j<i+(m>>1) do 66 begin 67 u:=a[j]; v:=mult(a[j+(m>>1)],cur[j-i]); 68 a[j]:=jia(u,v,1); 69 a[j+(m>>1)]:=jia(u,v,-1); 70 inc(j); 71 end; 72 i:=i+m; 73 end; 74 m:=m<<1; 75 end; 76 end; 77 78 begin 79 assign(input,'uoj34.in'); reset(input); 80 assign(output,'uoj34.out'); rewrite(output); 81 readln(n1,n2); 82 inc(n1); inc(n2); 83 for i:=0 to n1-1 do 84 begin 85 read(x); a[i].x:=x; 86 end; 87 for i:=0 to n2-1 do 88 begin 89 read(x); b[i].x:=x; 90 end; 91 n:=max(n1,n2); 92 m:=1; 93 while m<=(n<<1) do m:=m<<1; 94 fft(a,m,1); fft(b,m,1); 95 for i:=0 to m do a[i]:=mult(a[i],b[i]); 96 fft(a,m,-1); 97 for i:=0 to n1+n2-2 do write(round(a[i].x/m),' '); 98 close(input); 99 close(output); 100 end.