1到n打乱，求交换次数

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B - Old Sorting

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit  Status

Description

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

3

4

4 2 3 1

4

4 3 2 1

4

1 2 3 4

Sample Output

Case 1: 1

Case 2: 2

Case 3: 0

代码：

#include <stdio.h>
#include <stdlib.h>
#include <iostream>

using namespace std;

int a[105];
int main() {
	int cas;
	scanf("%d", &cas);
	for (int t = 1; t <= cas; t ++) {
		int n;
		int sum = 0;
		scanf("%d", &n);
		for (int i = 1; i <= n; i ++)
			scanf("%d", &a[i]);
		for(int i = 1; i <= n; i ++) {
			if(a[i] == i) continue;
			for(int j = i; j <= n; j ++) {
				if(a[j] == i) {
					a[j] = a[i];
					a[i] = i;
					sum ++;
					break;
				}
			}
		}
		printf("Case %d: %d\n",t, sum);
	}
	return 0;
}