### scott_wu's blog

By scott_wu, history, 14 months ago, Hey all,

I've posted my screencast and commentary of yesterday's Kickstart 2020 round B. Check it out at https://www.youtube.com/watch?v=AP74zQ0ZmRM

BTW, stay tuned for the return of Lockout streams in the coming weeks! :)  Comments (2)
 » Auto comment: topic has been updated by scott_wu (previous revision, new revision, compare).
 » Only scott_wu can choose overcomplicated solutions and still finish first :PHere's some simpler alternative solutions (which you alluded to in the analysis):problem B: - chosen solution: binary searching - simpler solution: going backwards & rounding D down to the nearest multiple of x[i]  ll n, d; cin >> n >> d; vector v(n); for (ll &i: v) cin >> i; for (int i = n - 1; i >= 0; --i) { d -= d % v[i]; } cout << d << "\n"; problem C: - chosen solution: something with a 2D array of pairs, unclear exactly how many of those values are used - simpler solution: iterate through string while maintaining a stack, latest value should be product of all loops that contain the current position typedef modnum mint; // https://github.com/ecnerwala/cp-book/blob/82a039f71e105737a1591caf609f4d1e0b9532ba/src/modnum.hpp string s; cin >> s; vector mult({1}); mint x = 0, y = 0; for (auto c: s) { if (c >= '0' && c <= '9') { mult.push_back(mult.back() * (c - '0')); } else if (c == '(') { } else if (c == ')') { mult.pop_back(); } else { mint dx = 0, dy = 0; if (c == 'E') ++dy; else if (c == 'W') --dy; else if (c == 'N') --dx; else ++dx; x += dx * mult.back(); y += dy * mult.back(); } } cout << int(y) + 1 << " " << int(x) + 1 << "\n";