数组相关题目：交换次数最小使得数组有序分布

Given an array containing a permutation of 1 to n, you have to find the minimum number of swaps to sort the array in ascending order. A swap means, you can exchange any two elements of the array.

For example, let n = 4, and the array be 4 2 3 1, then you can sort it in ascending order in just 1 swaps (by swapping 4 and 1).

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains two lines, the first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers separated by spaces. You may assume that the array will always contain a permutation of 1 to n.

Output

For each case, print the case number and the minimum number of swaps required to sort the array in ascending order.

Sample Input

Sample Input	Output for Sample Input
3 4 4 2 3 1 4 4 3 2 1 4 1 2 3 4	Case 1: 1 Case 2: 2 Case 3: 0

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 100+10;
int num[maxn];
bool vis[maxn];
int n;
#define REP(_,a,b) for(int _ = (a); _ <= (b); _++)

int main(){

    int ncase,T = 1;
    cin >> ncase;
    while(ncase--) {
        scanf("%d",&n);
        memset(vis,0,sizeof vis);
        int ans = 0;
        for(int i = 1; i <= n; i++) scanf("%d",&num[i]);
        for(int i = 1; i <= n; i++) {
            if(i==num[i]) continue;
            int idx;
            for(int j = i+1; j <= n; j++) {
                if(num[j]==i) {
                    idx = j;
                    break;
                }
            }

            num[idx] = num[i];

            ans++;

        }
        printf("Case %d: %d\n",T++,ans);
    }
    return 0;
}