Day 19 B. Flip the Bits

Problem:
There is a binary string a of length n. In one operation, you can select any prefix of a with an equal number of 0 and 1 symbols. Then all symbols in the prefix are inverted: each 0 becomes 1 and each 1 becomes 0.

For example, suppose a=0111010000.

In the first operation, we can select the prefix of length 8 since it has four 0’s and four 1’s: [01110100]00→[10001011]00.
In the second operation, we can select the prefix of length 2 since it has one 0 and one 1: [10]00101100→[01]00101100.
It is illegal to select the prefix of length 4 for the third operation, because it has three 0’s and one 1.
Can you transform the string a into the string b using some finite number of operations (possibly, none)?

Input
The first line contains a single integer t (1≤t≤10^4) — the number of test cases.

The first line of each test case contains a single integer n (1≤n≤3⋅10^5) — the length of the strings a and b.

The following two lines contain strings a and b of length n, consisting of symbols 0 and 1.

The sum of n across all test cases does not exceed 3⋅10^5.

Output
For each test case, output “YES” if it is possible to transform a into b, or “NO” if it is impossible. You can print each letter in any case (upper or lower).

Example
input
5
10
0111010000
0100101100
4
0000
0000
3
001
000
12
010101010101
100110011010
6
000111
110100
output
YES
YES
NO
YES
NO

Note
The first test case is shown in the statement.

In the second test case, we transform a into b by using zero operations.

In the third test case, there is no legal operation, so it is impossible to transform a into b.

In the fourth test case, here is one such transformation:

Select the length 2 prefix to get 100101010101.
Select the length 12 prefix to get 011010101010.
Select the length 8 prefix to get 100101011010.
Select the length 4 prefix to get 011001011010.
Select the length 6 prefix to get 100110011010.
In the fifth test case, the only legal operation is to transform a into 111000. From there, the only legal operation is to return to the string we started with, so we cannot transform a into b.

题目大致意思为：给你两串由01组成的字符串 你可以进行一些适当的操作 让第一串字符串变为第二串字符串 操作原则为 每次可以将0转变成1 将1转变成0 转变时只能将前缀转变 并且你得选择相同数量的0和1进行转换

#include<iostream>
#include<string.h>
#include<vector>
using namespace std;
int f[300005];
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		int n;
		cin >> n;
		for (int i = 0; i < n; i++)
		{
			f[i] = 0;
		}
		string sa, sb;
		cin >> sa >> sb;
		int xx = 0, yy = 0;
		for (int i = 0; i < sa.size(); i++) 
		{
			if (sa[i] == '0')
			{ 
				xx++; 
			}
			if (sa[i] == '1')
			{
				yy++;
			}
			if (xx == yy)
			{
				f[i] = 1;
			}
		}
		int change = 0;
		bool book = false;
		for (int i = sa.size() - 1; i >= 0; i--)
		{
			if (((sa[i] - '0') + change) % 2 != sb[i] - '0')
			{
				if (f[i])
				{
					change++;
				}
				else 
				{
					book = true;
					break;
				}
			}
		}
		if (!book)
		{
			cout << "YES" << endl;
		}
		else
		{
			cout << "NO" << endl;
		}
	}
	return 0;
}