Given any permutation（排列） of the numbers {0, 1, 2,..., N−1}, it is easy to sort them in increasing order. But what if ​​Swap(0, *)​​ is the ONLY operation that is allowed to use? For example, to sort {4, 0, 2, 1, 3} we may apply the swap operations in the following way:

Swap(0, 1) => {4, 1, 2, 0, 3}
Swap(0, 3) => {4, 1, 2, 3, 0}
Swap(0, 4) => {0, 1, 2, 3, 4}

Now you are asked to find the minimum number of swaps need to sort the given permutation of the first N nonnegative integers.

Input Specification:

Each input file contains one test case, which gives a positive N (≤105) followed by a permutation sequence of {0, 1, ..., N−1}. All the numbers in a line are separated by a space.

Output Specification:

For each case, simply print in a line the minimum number of swaps need to sort the given permutation.

Sample Input:

10
3 5 7 2 6 4 9 0 8 1

Sample Output:

9

参考代码：

#include<cstdio>
#include<algorithm>

using namespace std;

const int maxn = 100010;
int A[maxn];

int main(){
    int n;
    scanf("%d", &n);
    int left = n - 1;
    int num;
    for (int i = 0; i < n; ++i) {
        scanf("%d", &num);
        A[num] = i;
        if(num == i && num != 0) left--;//如果除0外，又在本位的数，left减1
    }

    int k = 1;//存放除0外不在本位的最小数
    int ans = 0;//记录交换次数
    while(left > 0){//只要还有数不在本位
        if(A[0] == 0){//如果0在本位，就寻找一个不在本位的数与0交换
            while(k < n){
                if(A[k] != k){          //找一个当前不在本位的数k
                    swap(A[0], A[k]);   //将k与0交换
                    ans++;
                    break;
                }
                k++;
            }
        }
        while(A[0] != 0){//只要0不再本位，就讲0与当前位置交换
            swap(A[0], A[A[0]]);        //将0与pos[0]交换
            ans++;                      //交换次数加一
            left--;
        }
    }
    printf("%d\n", ans);
    return 0;
}