B. Mister B and PR Shifts
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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.
Examples
input

3
1 2 3

output

0 0

input

3
2 3 1

output

0 1

input

3
3 2 1

output

2 1

解法：想了2天，蒟蒻还是没想到怎么做orz,只能求助网上的大神了。然后==会了一种乱搞的做法。
题意：有n个数，从1到n乱序排列，定义这n个数的秩序值为∑abs(a[i]-i) (1<=i<=n)， 你每次将这个数组向右循环移位p次，问p等于多少时，这n个数的秩序值最小？

记录下每个数是在目标位置的左边还是右边，存下所有在目标左边的数，
cur[i]表示初始数组中有多少个数在它目标左边第i个位置上
->注意要存下一开始在目标左边（包括在目标上）的个数L，以及目标右边的数的个数r
之后直接模拟右移，对于第i次右移，L = L-cur[i-1], r = r+cur[i-1]，即当次位移刚好有cur[i-1]个数原本在目标位置左边及目标上跑到了目标位置的右边，
每次的秩序值便是初始秩序值-L+r+特判最后一个数移到第一个数所产生的影响
但注意计算之后L还要+1，r还要-1因为最后一个数跑到了第一个
复杂度O(n)

思路来自：http://blog.csdn.net/jaihk662/article/details/73825315

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e6+7;
int n,cur[maxn*2],p[maxn],L,R,pos;
LL sum, ans;
int main()
{
    scanf("%d", &n);
    for(int i=1; i<=n; i++) scanf("%d", &p[i]);
    for(int i=1; i<=n; i++){
        if(p[i]>=i){
            L++;
            cur[p[i]-i]++;
        }
        else{
            R++;
        }
        sum += abs(i-p[i]);
    }
    ans = sum;
    for(int k = 0; k < n; k++){
        L -= cur[k];
        R += cur[k];
        sum = sum - L + R - abs(p[n-k]-(n+1)) + p[n-k] - 1;
        cur[p[n-k]+k]++;
        L++,R--;
        if(sum < ans){
            ans = sum, pos = k+1;
        }
    }
    printf("%lld %d\n", ans, pos);
    return 0;
}