Problem Description

You are given a permutation p1,p2,…,pN consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero):

Operation: Swap two adjacent elements in the permutation.

You want to have pi≠i for all 1≤i≤N. Find the minimum required number of operations to achieve this.

Constraints

2≤N≤105

p1,p2,..,pN is a permutation of 1,2,..,N.

Input

The input is given from Standard Input in the following format:

N
p1 p2 .. pN

Output

Print the minimum required number of operations

Example

Sample Input 1

5
1 4 3 5 2

Sample Output 1

2
Swap 1 and 4, then swap 1 and 3. p is now 4,3,1,5,2 and satisfies the condition. This is the minimum possible number, so the answer is 2.

Sample Input 2

2
1 2

Sample Output 2

1
Swapping 1 and 2 satisfies the condition.

Sample Input 3

2
2 1

Sample Output 3

0
The condition is already satisfied initially.

Sample Input 4

9
1 2 4 9 5 8 7 3 6

Sample Output 4

3

题意：给出一个长度为 n 的序列 a，序列中的数分别为 1~n，现在可以对序列中相邻元素进行交换，问最少交换多少次能使得 ai≠i

思路：从前向后扫一遍记录是否需要交换即可

Source Program

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 100000+5;
const int dx[] = {-1,1,0,0,-1,-1,1,1};
const int dy[] = {0,0,-1,1,-1,1,-1,1};
using namespace std;int a[N];
int main() {int n;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);int res=0;for(int i=1;i<=n-1;i++){if(a[i]==i){swap(a[i],a[i+1]);res++;}}if(a[n]==n)res++;printf("%d\n",res);return 0;
}

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Derangement（AtCoder-3525）

Problem Description

Constraints

Input

Output

Example

Source Program

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