Разбор
Problem Author: | Ivan Balashov, Kostiantyn Denysov |
Prepared by: | Kostiantyn Denysov |
Editorial by: | Kostiantyn Denysov |
We can independently check whether it is possible to make the -coordinate equal to and whether it is possible to make the -coordinate equal to .
To check if it is possible to make the -coordinate equal to , we calculate the total number of letters and in the given string. Then, we brute force all possible counts of in the final string (denote this count as ) and determine the count of as . Finally, we check if the following equation holds for the -coordinate:
Time complexity:
//#pragma GCC optimize("Ofast", "unroll-loops") //#pragma GCC target("sse", "sse2", "sse3", "ssse3", "sse4") #include <bits/stdc++.h> #define all(a) a.begin(),a.end() #define len(a) (int)(a.size()) #define mp make_pair #define pb push_back #define fir first #define sec second #define fi first #define se second using namespace std; typedef pair<int, int> pii; typedef long long ll; typedef long double ld; template<typename T> bool umin(T &a, T b) { if (b < a) { a = b; return true; } return false; } template<typename T> bool umax(T &a, T b) { if (a < b) { a = b; return true; } return false; } #ifdef KIVI #define DEBUG for (bool _FLAG = true; _FLAG; _FLAG = false) #define LOG(...) print(#__VA_ARGS__" ::", __VA_ARGS__) << endl template <class ...Ts> auto &print(Ts ...ts) { return ((cerr << ts << " "), ...); } #else #define DEBUG while (false) #define LOG(...) #endif const int max_n = -1, inf = 1000111222; inline bool good (int n, int a, int b) { for (int i = 0; i <= n; i++) { ll now = i * 1ll * a - (n - i) * 1ll * b; if (now == 0) { return true; } } return false; } inline void test_case () { int n, N, S, E, W; cin >> n >> E >> W >> N >> S; int cnt[2] = {}; string s; cin >> s; for (char c : s) { if (c == 'N' || c == 'S') { ++cnt[0]; } else { ++cnt[1]; } } if (good(cnt[0], N, S) && good(cnt[1], E, W)) { cout << "Yes\n"; } else { cout << "No\n"; } } int main() { // freopen("input.txt", "r", stdin); // freopen("output.txt", "w", stdout); ios_base::sync_with_stdio(0); cin.tie(0); int t = 1; cin >> t; for (int test = 1; test <= t; test++) { test_case(); } }