AtCoder Regular Contest 082 D Derangement

AtCoder Regular Contest 082 D Derangement

与下标相同与下个交换就好了。。。。

Define a sequence of ’o’ and ’x’ of length N as follows: if pi ̸= i, the i-th symbol is ’o’, otherwise the i-th symbol is ’x’. Our objective is to change this sequence to ’ooo...ooo’.
• If there is a part ”ox” (or ”xo”) in the sequence, we can change it to ”oo” by swapping these two elements. (∵ If pi = x(x ̸= i) and pi+1 = i + 1 in the initial sequence, after the swap, both pi = i + 1 and pi+1 = x will be ’o’.) • If there is a part ”xx” in the sequence, we can change it to ”oo” by swapping these two elements. (∵ If pi = i(x ̸= i) and pi+1 = i + 1 in the initial sequence, after the swap, both pi = i + 1 and pi+1 = i will be ’o’.)
Thus, we should check the sequence from left to right, and if we find an ’x’ at the i-th position, we should swap i and i + 1 (unless i = N, in this case we should swap i and i−1).

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+10;
int a[N];

int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=1;i<=n;i++) cin>>a[i];
        int cnt=0;
        for(int i=1;i<=n;i++){
            if(a[i]==i){
                cnt++;
                swap(a[i],a[i+1]);
            }
        }
        cout<<cnt<<endl;
    }
    return 0;
}