$Description$
Given equal length $ a, b $ two sequences each time sequence may exchange a number of two adjacent minimum required number of exchanges that $ \ sum (a_i-b_i) ^ 2 $ minimized.
$Sol$
Swap meet certain sequence after sequence $ a $ large number of the first $ i $ and $ b $ in the sequence of the $ i $ large number corresponds to. Prove quite obviously did not say, Los ri explanations given some a pair then the number of discrete two sequences, the sequence of conversion $ b $ $ I $ of the large number of $ $ to position a $ I $ sequence where a large number. the question then transformed into , $ b $ exchange sequence how many times can turn into $ a $, and because the discrete $ a $ sequence is a single increase, in fact, $ 1,2,3 ....., n $. so this is only required when $ b $ sequence reverse to like you may be used to sort or merge tree array.
$Code$
#include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<algorithm> #define il inline #define Rg register #define go(i,a,b) for(Rg int i=a;i<=b;++i) #define yes(i,a,b) for(Rg int i=a;i>=b;--i) #define mem(a,b) memset(a,b,sizeof(a)) #define ll long long #define db double #define inf 2147483647 using namespace std; il int read() { Rg int x=0,y=1;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();} while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();} return x*y; } const int N=100010,v =99999997; int n,c[N],d[N],e[N]; ll as; struct node{int w,pos,dx;}a[N],b[N]; il bool cmp1(node x,node y){return x.w<y.w;} il bool cmp2(node x,node y){return x.pos<y.pos;} il void lsh() { sort(a+1,a+n+1,cmp1); sort(b+1,b+n+1,cmp1); go(i,1,n)c[i]=a[i].pos; go(i,1,n)b[i].dx=i; sort(b+1,b+n+1,cmp2); go(i,1,n)d[i]=c[b[i].dx]; } il void gb(int l,int r) { if(l==r)return; Rg int mid=(l+r)>>1,i=l,j=mid+1; gb(l,mid);gb(mid+1,r); go(k,l,r) { if((d[i]>d[j]&&j<=r)||i>mid)as=(as+(mid-i+1))%mod,e[k]=d[j++]; else e[k]=d[i++]; } go(k,l,r)d[k]=e[k]; } int main() { n=read(); go(i,1,n)a[i].w=read(),a[i].pos=i; go(i,1,n)b[i].w=read(),b[i].pos=i; lsh(); gb(1,n); printf("%lld\n",as); return 0; }