Derangement

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.

把数组从头到尾过一遍再从尾到头过一遍就好啦，不过很容易忘记最后一个数呢23333（用C也能做呢）

#include 
#include 
#include 
using namespace std;
int main(int argc, char *argv[])
{
	int n,i,f,g=1,sum=0;
	int a[100000];
	int b[100000];
	memset(b,0,sizeof(b));
	scanf("%d",&n);
	for(i=0;i=0;i--)
	{
		if(a[i]==i+1)
		{
			int t;
			t=a[i];
			a[i]=a[i+1];
			a[i+1]=t;
			b[i]--;
		}
	}
	for(i=0;i