文章目录
一、B. Even Array
总结
一、B. Even Array
本题链接：B. Even Array
题目：
B. Even Array
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given an array a[0…n−1] of length n which consists of non-negative integers. Note that array indices start from zero.
An array is called good if the parity of each index matches the parity of the element at that index. More formally, an array is good if for all i (0≤i≤n−1) the equality imod2=a[i]mod2 holds, where xmod2 is the remainder of dividing x by 2.
For example, the arrays [0,5,2,1] and [0,17,0,3] are good, and the array [2,4,6,7] is bad, because for i=1, the parities of i and a[i] are different: imod2=1mod2=1, but a[i]mod2=4mod2=0.
In one move, you can take any two elements of the array and swap them (these elements are not necessarily adjacent).
Find the minimum number of moves in which you can make the array a good, or say that this is not possible.
Input
The first line contains a single integer t (1≤t≤1000) — the number of test cases in the test. Then t test cases follow.
Each test case starts with a line containing an integer n (1≤n≤40) — the length of the array a.
The next line contains n integersa0,a1,…,an−1 (0≤ai≤1000) — the initial array.
Output
For each test case, output a single integer — the minimum number of moves to make the given array a good, or -1 if this is not possible.
Example
input
4
4
3 2 7 6
3
3 2 6
1
7
7
4 9 2 1 18 3 0
output
2
1
-1
0
Note
In the first test case, in the first move, you can swap the elements with indices 0 and 1, and in the second move, you can swap the elements with indices 2 and 3.
In the second test case, in the first move, you need to swap the elements with indices 0 and 1.
In the third test case, you cannot make the array good.
本博客给出本题截图：
题意：要求数组从0开始，奇数的索引必须元素为奇数，偶数的索引必须元素为偶数，问对于一个数组，每次可以交换任意两个位置上的元素，问最少交换几次可以使数组满足上述需求，如果不可以，输出-1
AC代码
#include <iostream> using namespace std; const int N = 1010; int a[N]; int main() { int t; cin >> t; while (t -- ) { int n, odd = 0, even = 0; cin >> n; for (int i = 0; i < n; i ++ ) { cin >> a[i]; if (i % 2 == 0 && a[i] % 2 != 0) odd ++; else if (i % 2 != 0 && a[i] % 2 == 0) even ++; } if (odd != even) puts("-1"); else cout << odd << endl; } return 0; }
总结
水题，不解释
