https://atcoder.jp/contests/abc421/tasks/abc421_d
按相對位移去處理兩個人的移動可以簡化模擬難度

#include <bits/stdc++.h>
#define nmf(i, s, e) for (int i = s; i <= e; i++)
#define ref(i, s, e) for (int i = s; i >= e; i--)
using namespace std;
typedef long long LL;
typedef unsigned long long uLL;
int main()
{
    pair<int, int> mp[4];
    mp[0] = {0, -1}, mp[1] = {0, 1}, mp[2] = {-1, 0}, mp[3] = {1, 0};
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    LL rt, ct, ra, ca;
    cin >> rt >> ct >> ra >> ca;
    ra -= rt;
    ca -= ct;
    rt = ct = 0;
    LL n, m, l;
    cin >> n >> m >> l;
    deque<pair<int, LL>> a, b;
    nmf(i, 1, m)
    {
        char c;
        LL cnt;
        cin >> c >> cnt;
        a.push_back({(c == 'L' ? 0 : c == 'R' ? 1
                                 : c == 'U'   ? 2
                                              : 3),
                     cnt});
    }
    nmf(i, 1, l)
    {
        char c;
        LL cnt;
        cin >> c >> cnt;
        b.push_back({(c == 'L' ? 0 : c == 'R' ? 1
                                 : c == 'U'   ? 2
                                              : 3),
                     cnt});
    }
    LL ans = 0;
    while (!a.empty() || !b.empty())
    {
        int dt = a.front().first;
        LL la = a.front().second;
        int da = b.front().first;
        LL lb = b.front().second;
        a.pop_front();
        b.pop_front();
        LL len = min(la, lb);
        if (la > lb)
        {
            a.push_front({dt, la - len});
        }
        else if (la < lb)
        {
            b.push_front({da, lb - len});
        }
        LL d1 = mp[da].first - mp[dt].first;
        LL d2 = mp[da].second - mp[dt].second;
        LL step = -1;
        if (rt == ra && ct == ca)
        {
            if (dt == da)
                ans += len;
        }
        else if (ra || ca)
        {
            if (d1 != 0)
                step = -ra / d1;
            if (d2 != 0)
                step = -ca / d2;
            if (step > 0 && step <= len && d1 * step == -ra && d2 * step == -ca)
                ans++;
        }
        ra += d1 * len;
        ca += d2 * len;
    }
    cout << ans << endl;
    return 0;
}