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+;
 int a[N];

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