Some time ago Mister B detected a strange signal from the space, which he started to study.
After some transformation the signal turned out to be a permutation p of length n or its cyclic shift. For the further investigation Mister B need some basis, that's why he decided to choose cyclic shift of this permutation which has the minimum possible deviation.
Let's define the deviation of a permutation p as .
Find a cyclic shift of permutation p with minimum possible deviation. If there are multiple solutions, print any of them.
Let's denote id k (0 ≤ k < n) of a cyclic shift of permutation p as the number of right shifts needed to reach this shift, for example:
k = 0: shift p1, p2, ... pn,
k = 1: shift pn, p1, ... pn - 1,
...,
k = n - 1: shift p2, p3, ... pn, p1.
Input
First line contains single integer n (2 ≤ n ≤ 106) — the length of the permutation.
The second line contains n space-separated integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the elements of the permutation. It is guaranteed that all elements are distinct.
Output
Print two integers: the minimum deviation of cyclic shifts of permutation p and the id of such shift. If there are multiple solutions, print any of them.
题解
对于一个数\(i\)，越往第\(i\)个这个位置靠近，差值越小，相反差值越大。
于是我们考虑记录转变的时间，在不变的时候，一直可以保证它的单调性，每次往答案里\(+1\)或者\(-1\)就可以，然后O(n)去更新，最后记得从n到1时的变化要单独更新。
更详细的看代码把。
代码
#include
#include
#include
#include
using namespace std;
const int maxn=1e6+10;
int cnt[maxn];
int a[maxn],n;
int abs(int x){return x>0?x:-x;}
int main(){
ios::sync_with_stdio(false);
while(cin>>n){
int add=0,sub=0,b=0;
long long tmp=0,res;
memset(cnt,0,sizeof(cnt));
for(int i=1;i<=n;i++){
cin>>a[i];
tmp+=abs(a[i]-i);
cnt[(a[i]-i+n)%n]++;//记录第几次移动后单调性会变
if(a[i]>i)sub++;
else add++;
}
res=tmp;
for(int i=1;itmp){
res=tmp;
b=i;
}
}
cout<