4476: A. Make A Equal to B
内存限制:256 MB
时间限制:1 S
标准输入输出
题目类型:传统
评测方式:Special Judge
上传者:
提交:22
通过:16
题目描述
You are given two arrays a and b of n elements, each element is either 0 or 1.
You can make operations of 2 kinds.
You can make operations of 2 kinds.
- Pick an index i and change ai to 1−ai
- Rearrange the array a however you want.
输入格式
Each test contains multiple test cases. The first line contains a single integer t (1 <= t <= 400) — the number of test cases. Description of the test cases follows.
The first line of each test case contains a single integer n (1 <= n <= 100) — the length of the arrays a and b.
The second line of each test case contains n space-separated integers a1,a2,…,an (ai is 0 or 1), representing the array a.
The third line of each test case contains n space-separated integers b1,b2,…,bn (bi is 0 or 1), representing the array b.
The first line of each test case contains a single integer n (1 <= n <= 100) — the length of the arrays a and b.
The second line of each test case contains n space-separated integers a1,a2,…,an (ai is 0 or 1), representing the array a.
The third line of each test case contains n space-separated integers b1,b2,…,bn (bi is 0 or 1), representing the array b.
输出格式
For each test case, print the minimum number of operations required to make a equal to b.
输入样例 复制
5
3
1 0 1
0 0 1
4
1 1 0 0
0 1 1 1
2
1 1
1 1
4
1 0 0 1
0 1 1 0
1
0
1
输出样例 复制
1
2
0
1
1
数据范围与提示
In the first case, we need only one operation: change a1 to 1-ai. Now a = [0, 0] which is equal to b.
In the second case, the optimal way is to rearrange a to get the array [0, 1, 11. Now a = [0, 0, 1] which is equal to b.
In the second case, one of optimal ways would be to first change a3 to 1 - a3, then rearrange a.
In the third case, no operation is needed.
In the fourth case, the optimal way is to rearrange a to get the array [0, 1, 1, 0].
In the second case, the optimal way is to rearrange a to get the array [0, 1, 11. Now a = [0, 0, 1] which is equal to b.
In the second case, one of optimal ways would be to first change a3 to 1 - a3, then rearrange a.
In the third case, no operation is needed.
In the fourth case, the optimal way is to rearrange a to get the array [0, 1, 1, 0].