코딩테스트/백준

[골3] 1958 - LCS 3

#include #include using namespace std; #pragma region 입출력 속도향상 #define FAST_IO() \ {\ ios::sync_with_stdio(false);\ cin.tie(NULL); \ cout.tie(NULL); \ }\ #pragma endregion #define MAX 101 int dp[MAX][MAX][MAX]; int max(int a, int b, int c) { return max(a, max(b, c)); } int main() { FAST_IO(); string s1, s2, s3; cin >> s1 >> s2 >> s3; int sz1 = s1.length(), sz2 = s2.length(), sz3 = s3.length(); f..

[실4] 15489 - 파스칼의 삼각형

dp를 활용했다 #pragma region 입출력 속도향상 #define FAST_IO() \ {\ ios::sync_with_stdio(false);\ cin.tie(NULL); \ cout.tie(NULL); \ }\ #pragma endregion #include using namespace std; #define SIZE 30 int dp[SIZE + 1][SIZE + 1]; int main() { FAST_IO(); int r, c, w; cin >> r >> c >> w; dp[1][1] = 1; for (int i = 1; i

[실5] 16395 - 파스칼의 삼각형

#pragma region 입출력 속도향상 #define FAST_IO() \ {\ ios::sync_with_stdio(false);\ cin.tie(NULL); \ cout.tie(NULL); \ }\ #pragma endregion #include #include using namespace std; int main() { FAST_IO(); int n, k; cin >> n >> k; vector pascal(n); for (int i = 0; i < n; i++) { int val = 1; for (int k = 0; k

[골3] 10986 - 나머지 합 구하기

#pragma region 입출력 속도향상 #define FAST_IO() \ {\ ios::sync_with_stdio(false);\ cin.tie(NULL); \ cout.tie(NULL); \ }\ #pragma endregion #include #include using namespace std; int main() { FAST_IO(); int n, m; cin >> n >> m; vector v(n + 1), c(m); vector s(n + 1); long ans = 0; for (int i = 1; i > tmp; v[i] = (v[i - 1] + tmp) % m; ans += (v[i] == 0); c[v[i]]++; } for (const auto& val : c) ans += (va..

[골5] 1806 - 부분합

#include #include #include using namespace std; #define MAX 987654321 int main() { int n, r; cin >> n >> r; vector v(100001); for (int i = 0; i > v[i]; int s = 0, e = 0, res = MAX; int sum = v[0]; while (true) { if (sum >= r) { res = min(res, e - s + 1); sum -= v[s++]; } else { if (++e == n) break; sum += v[e]; } } cout

[실3] 3273 - 두 수의 합

#include #include #include using namespace std; int main() { int n; cin >> n; vector v(n); for (int i = 0; i > v[i]; sort(v.begin(), v.end()); int x; cin >> x; int s = 0, e = n - 1, cnt = 0; while (s < e) { int a = v[s], b = v[e]; int sum = a + b; if (sum == x) cnt++; if (sum < x) s++; else e--; } cout

[골5] 2294 - 동전2

#include #include #pragma region 입출력 속도향상 #define FAST_IO() \ {\ ios::sync_with_stdio(false);\ cin.tie(NULL); \ cout.tie(NULL); \ }\ #pragma endregion using namespace std; int main() { FAST_IO(); int n, k; cin >> n >> k; vector v(n), dp(k + 1); for (int i = 0; i > v[i]; for (int i = 1; i

[실3] 11659 - 구간 합 구하기 4

#include #include using namespace std; #pragma region 입출력 속도향상 #define FAST_IO() \ {\ ios::sync_with_stdio(false);\ cin.tie(NULL); \ cout.tie(NULL); \ }\ #pragma endregion int main() { FAST_IO(); int n, m; cin >> n >> m; vector v(n + 1), t(n + 1); // 입력 for (int i = 1; i > v[i]; // 구간합 구하기 for (int i = 1; i > s >> e; cout

[골5] 1106 - 호텔

#include #include #include using namespace std; #pragma region 입출력 속도향상 #define FAST_IO() \ {\ ios::sync_with_stdio(false);\ cin.tie(NULL); \ cout.tie(NULL); \ }\ #pragma endregion #define MAX 100000 int main() { FAST_IO(); int c, n; cin >> c >> n; vector v; for (int i = 0; i > a >> b; v.push_back({ a,b }); } int dp[MAX + 1] = { 0 }; for (int i = 0; i < v.size(); i++) ..

[골5] 12904 - A와 B

#include #include #include using namespace std; #pragma region 입출력 속도향상 #define FAST_IO() \ {\ ios::sync_with_stdio(false);\ cin.tie(NULL); \ cout.tie(NULL); \ }\ #pragma endregion int main() { FAST_IO(); string s, t; cin >> s >> t; while (t.length() != s.length()) { if (t.back() != 'A') { t.pop_back(); reverse(t.begin(), t.end()); } else t.pop_back(); } cout