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

題意：給出一個(gè)長(zhǎng)度為 n 的序列 a，序列中的數(shù)分別為?1~n，現(xiàn)在可以對(duì)序列中相鄰元素進(jìn)行交換，問(wèn)最少交換多少次能使得 ai≠i

思路：從前向后掃一遍記錄是否需要交換即可

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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