AcWing 1995. Meet and greet (simulation)

【 Title Description 】
as everyone knows , Cows are very socially polite animals ： Whenever two cows meet after they separate , They all use friendly “ Moo ” Greet each other .
Bessie the cow and her friend ash are walking along a long road on Farmer John's farm .
We can think of this path as a one-dimensional number axis .
Bessie and ashy both start from the origin , At the same speed （$1$ Unit distance / Unit time ） Walk for a while .
Please describe the movement of each cow , Determine how many times they greet each other .
Bessie and ash can stop moving at different time points , And they don't move longer than $1000000$ Company .

【 Input format 】
The first line contains two integers $B$ and $E$.
Next $B$ That's ok , Describes Bessie's movement . Each line contains a positive integer followed by a character $L$ or $R$, It means Bessie is in a certain direction （ Left or right ） Moved several units of distance .
Next $E$ That's ok , Describes ash's movement . Each line contains a positive integer followed by a character $L$ or $R$, It means Ashley is in a certain direction （ Left or right ） Moved several units of distance .

【 Output format 】
Output the number of times they greet each other .
Be careful , When initially in the initial position , They don't say hello .

【 Data range 】
$1\le B,E\le 50000$

【 sample input 】

4 5
3 L
5 R
1 L
2 R
4 R
1 L
3 L
4 R
2 L

【 sample output 】

3

【 Sample explanation 】
Bessie and ash are in time $7,9,13$ Meet and say hello .

【 analysis 】

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At the beginning, both cows were at the origin , Therefore, the point parity of two cows is the same , According to the distance that two cows move one unit per unit of time, it can be seen that when two cows move at the same time, they must maintain the same parity of their positions ; When a cow stops moving , It must stop at the whole point , So whenever two cows meet, they must meet on the whole point , That is, at the end of a whole unit of time , Instead of meeting in the middle .

Then we can simulate and calculate the position of two cows in each unit time , Then enumerate each moment $i$, If $i$ At all times, the two cows are in the same position and $i-1$ The position of the two cows is different at all times , That means two cows met , Add one... To the answer .

---

【 Code 】

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 1000010;
int pos1[N], pos2[N];
int n, m, res;

int main()
{
    
    cin >> n >> m;
    int t = 0;
    while (n--)
    {
    
        char d;
        int x, v = 1;
        cin >> x >> d;
        if (d == 'L') v = -1;
        while (x--) t++, pos1[t] = pos1[t - 1] + v;
    }
    while (t < N - 1) t++, pos1[t] = pos1[t - 1];
    t = 0;
    while (m--)
    {
    
        char d;
        int x, v = 1;
        cin >> x >> d;
        if (d == 'L') v = -1;
        while (x--) t++, pos2[t] = pos2[t - 1] + v;
    }
    while (t < N - 1) t++, pos2[t] = pos2[t - 1];
    for (int i = 1; i < N; i++)
        if (pos1[i] == pos2[i] && pos1[i - 1] != pos2[i - 1]) res++;
    cout << res << endl;
    return 0;
}