description

solution

可以把點投射到以i座標為x座標，以a[i]為y座標的二維平面，考慮維護兩條斜率為1直線，點到直線豎直距離和即為答案，分別為維護多少個點在直線上，用兩個桶維護即可，座標分別為到y=x豎直距離和到(n,0)曼哈頓距離。

code

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LF double
#define LL long long
#define ULL unsigned LL
#define fo(i,j,k) for(LL i=j;i<=k;i++)
 

#define fd(i,j,k) for(LL i=j;i>=k;i--)
#define fr(i,j) for(LL i=begin[j];i;i=next[i])
using namespace std;
LL const mn=4*1e6+2,inf=1e9+7;
LL n,a[mn],cnt[mn],cn2[mn];
LL read(){
    char ch=getchar();
    while((ch!='-')&&((ch<'0')||(ch>'9')))ch=getchar();
    LL op,v=0;
    if(ch=='-'
 
)op=-1;
    else op=1,v=ch-'0';
    ch=getchar();
    while((ch>='0')&&(ch<='9'))v=v*10+ch-'0',ch=getchar();
    return op*v;
}
int main(){
    freopen("d.in","r",stdin);
    freopen("d.out","w",stdout);
    n=read();
    fo(i,1,n)a[i]=read();
    LL up=0,tmp=0;
    fo(i,1,n){
        tmp+=abs
 
(a[i]-i);
        if(a[i]>i)cnt[a[i]-i]++;
        up+=a[i]>i;
    }
    LL ans=tmp,up2=0;
    fd(i,n,2){
        tmp+=i-1-up*2;
        up-=cnt[n-i+1];
        tmp+=a[i]+1-abs(a[i]-n-1);

        cn2[a[i]+n-i]++;
        up2++;

        tmp+=n-i+1-up2*2;
        up2-=cn2[n-i+1];
        ans=min(ans,tmp);
    }
    printf("%lld",ans);
    return 0;
}