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| // Alchemy | |
| // Solution by Jacob Plachta | |
| using namespace std; | |
| template<typename T> T Abs(T x) { return(x < 0 ? -x : x); } | |
| template<typename T> T Sqr(T x) { return(x * x); } | |
| string plural(string s) { return(Sz(s) && s[Sz(s) - 1] == 'x' ? s + "en" : s + "s"); } | |
| const int INF = (int)1e9; | |
| const LD EPS = 1e-12; | |
| const LD PI = acos(-1.0); | |
| bool Read(int& x) { | |
| char c, r = 0, n = 0; | |
| x = 0; | |
| for (;;) { | |
| c = GETCHAR(); | |
| if ((c < 0) && (!r)) | |
| return(0); | |
| if ((c == '-') && (!r)) | |
| n = 1; | |
| else if ((c >= '0') && (c <= '9')) | |
| x = x * 10 + c - '0', r = 1; | |
| else if (r) | |
| break; | |
| } | |
| if (n) | |
| x = -x; | |
| return(1); | |
| } | |
| int N; | |
| char P[LIM]; | |
| bool ProcessCase() | |
| { | |
| // input | |
| Read(N); | |
| scanf("%s", P); | |
| // count difference between counts of As and Bs | |
| int i, c = 0; | |
| Fox(i, N) | |
| c += (P[i] == 'A' ? 1 : -1); | |
| // absolute difference must be equal to 1 | |
| return(Abs(c) == 1); | |
| } | |
| int main() { | |
| int T, t; | |
| Read(T); | |
| Fox1(t, T) | |
| printf("Case #%d: %c\n", t, ProcessCase() ? 'Y' : 'N'); | |
| return(0); | |
| } |