题目
解题思路:
模拟
我们先求出最初S的健美值,然后不旋转S,只对Si的下标进行操作
如果往左移动,当S[i]>i时,移动一次,健美值-1,当S[i]<=i时,移动一次则健美值+1.
Accepted code:
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define fo(i,a,b) for(i=a;i<=b;i++)
using namespace std;
const int maxn=4e6+7;
typedef long long ll;
ll i,j,k,l,t,n,m,ans,an;
ll a[maxn],bz[maxn],bb,cc;
void read(ll &f) {
f=0; char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) f=(f<<1)+(f<<3)+c-48,c=getchar();
return;
}
int main(){
read(n);
fo(i,1,n){
read(a[i]);
if(a[i]>i)ans+=a[i]-i,bz[min(n-i+a[i],a[i]-i)]++,bb++;
else ans+=i-a[i],cc++;
}
an=ans;
fo(i,1,n){
ans+=cc-bb;
ans-=n+1-a[n-i+1];
ans+=a[n-i+1]-1;
cc--;
if(a[n-i+1]>1)bb++,bz[i+a[n-i+1]-1]++;
else cc++;
bb-=bz[i],cc+=bz[i];
an=min(an,ans);
}
printf("%lld\n",an);
}