题意：
思路：可以看出题目所要最小化的是这样一个形式：
拆出每一项之后发现会变化的项只有sigma a[i]*b[i+t]与c^2,c*(a[i]-b[i])
c可以在外层枚举，剩下的只有sigma a[i]*b[i+t] (i=0..n-1)
因为FFT只能解决simga a[i]*b[n-i]
所以我们可以把a翻转，这样就化成了如上的形式
c[n+t+1]=a[n-i+1]*b[i+t] (i=0..n-1)
取出最小（大）的c[i]，与外层枚举的c共同求出答案，取最小值
FFT模板，值得一背
1 type cp=record 2 x,y:extended; 3 end; 4 arr=array[0..510000]of cp; 5 var c,d,cur:arr; 6 a,b:array[0..510000]of longint; 7 n,i,m:longint; 8 ans,sum,s1,s2:int64; 9 10 procedure swap(var x,y:cp); 11 var t:cp; 12 begin 13 t:=x; x:=y; y:=t; 14 end; 15 16 function jia(a,b:cp;f:longint):cp; 17 begin 18 if f=-1 then 19 begin 20 b.x:=-b.x; b.y:=-b.y; 21 end; 22 jia.x:=a.x+b.x; 23 jia.y:=a.y+b.y; 24 end; 25 26 function mult(a,b:cp):cp; 27 begin 28 mult.x:=a.x*b.x-a.y*b.y; 29 mult.y:=a.x*b.y+a.y*b.x; 30 end; 31 32 function min(x,y:int64):int64; 33 begin 34 if x<y then exit(x); 35 exit(y); 36 end; 37 38 function max(x,y:int64):int64; 39 begin 40 if x>y then exit(x); 41 exit(y); 42 end; 43 44 procedure fft(var a:arr;n,f:longint); 45 var i,j,k,m:longint; 46 w,u,v:cp; 47 begin 48 i:=n>>1; j:=1; 49 while j<n do 50 begin 51 if i<j then swap(a[i],a[j]); 52 k:=n>>1; 53 while k and i>0 do 54 begin 55 i:=i xor k; 56 k:=k>>1; 57 end; 58 i:=i xor k; 59 inc(j); 60 end; 61 m:=2; 62 while m<=n do 63 begin 64 w.x:=cos(2*pi*f/m); w.y:=sin(2*pi*f/m); 65 cur[0].x:=1; cur[0].y:=0; 66 for i:=1 to m-1 do cur[i]:=mult(cur[i-1],w); 67 i:=0; 68 while i<n do 69 begin 70 j:=i; 71 while j<i+(m>>1) do 72 begin 73 u:=a[j]; v:=mult(a[j+(m>>1)],cur[j-i]); 74 a[j]:=jia(u,v,1); 75 a[j+(m>>1)]:=jia(u,v,-1); 76 inc(j); 77 end; 78 i:=i+m; 79 end; 80 m:=m<<1; 81 end; 82 end; 83 84 procedure solve; 85 var i,j,len:longint; 86 begin 87 for i:=0 to n-1 do c[i].x:=a[i]; 88 for i:=0 to n-1 do d[i].x:=b[i]; 89 i:=0; j:=n-1; 90 while i<=j do 91 begin 92 swap(d[i],d[j]); 93 inc(i); dec(j); 94 end; 95 len:=1; 96 while len<=n+n do len:=len<<1; 97 for i:=n to len-1 do 98 begin 99 c[i].x:=0; d[i].x:=0; 100 end; 101 for i:=0 to len-1 do 102 begin 103 c[i].y:=0; d[i].y:=0; 104 end; 105 fft(c,len,1); 106 fft(d,len,1); 107 for i:=0 to len-1 do c[i]:=mult(c[i],d[i]); 108 fft(c,len,-1); 109 for i:=0 to len-1 do c[i].x:=trunc(c[i].x/len+0.5); 110 for i:=n to len-1 do c[i mod n].x:=c[i mod n].x+c[i].x; 111 sum:=-(1<<62); 112 for i:=0 to n-1 do sum:=max(sum,round(c[i].x)); 113 sum:=-sum*2; 114 end; 115 116 begin 117 assign(input,'bzoj4827.in'); reset(input); 118 assign(output,'bzoj4827.out'); rewrite(output); 119 readln(n,m); 120 for i:=0 to n-1 do read(a[i]); 121 for i:=0 to n-1 do read(b[i]); 122 ans:=1<<62; 123 solve; 124 for i:=0 to n-1 do 125 begin 126 s1:=s1+a[i]-b[i]; 127 s2:=s2+a[i]*a[i]+b[i]*b[i]; 128 end; 129 for i:=-m to m do ans:=min(ans,int64(n)*i*i+2*s1*i+s2+sum); 130 writeln(ans); 131 close(input); 132 close(output); 133 end.