随笔分类 - AcWing
摘要：#include <iostream> using namespace std; const int N = 310; int n; int s[N]; int f[N][N]; int main() { cin >> n; for (int i = 1; i <= n; ++ i) cin >>
阅读全文
摘要：#include <iostream> using namespace std; const int N = 1010; int n, m; char a[N], b[N]; int f[N][N]; int main() { cin >> n >> m; cin >> a + 1 >> b + 1
阅读全文
摘要：#include <iostream> using namespace std; const int N = 1010; int n; int a[N], f[N]; int main() { cin >> n; for (int i = 1; i <= n; ++ i) cin >> a[i];
阅读全文
摘要：#include <iostream> using namespace std; const int N = 550, INF = 1e9; int n; int a[N][N], f[N][N]; int main() { cin >> n; for (int i = 1; i <= n; ++
阅读全文
摘要：#include <iostream> using namespace std; const int N = 110; int v[N][N], w[N][N]; int s[N], f[N]; int n, m; int main() { cin >> n >> m; for (int i = 1
阅读全文
摘要：// 朴素解法 #include <iostream> using namespace std; const int N = 110; int n, m; int s[N], v[N], w[N]; int f[N][N]; int main() { cin >> n >> m; for (int
阅读全文
摘要：#include <iostream> using namespace std; const int N = 1010; int n, m; int v[N], w[N]; int f[N]; int main() { cin >> n >> m; for (int i = 1; i <= n; +
阅读全文
摘要：#include <iostream> using namespace std; const int N = 1010; int n, m; int v[N], w[N]; int f[N]; int main() { cin >> n >> m; for (int i = 1; i <= n; +
阅读全文
摘要：#include <iostream> #include <vector> #include <algorithm> using namespace std; vector<int> div(vector<int> A, int b, int &r) { vector<int> C; r = 0;
阅读全文
摘要：#include <iostream> #include <vector> using namespace std; vector<int> mul(vector<int> A, int b) { vector<int> C; int t = 0; for (int i = 0; i < A.siz
阅读全文
摘要：#include <iostream> #include <vector> using namespace std; bool cmp(vector<int> A, vector<int> B) { if (A.size() != B.size()) return A.size() > B.size
阅读全文
摘要：#include <iostream> #include <vector> using namespace std; vector<int> add(vector<int> A, vector<int> B) { vector<int> C; int t = 0; for (int i = 0; i
阅读全文
摘要：#include <cstdio> #include <cmath> int main() { double x; scanf("%lf", &x); double l = -10000, r = 10000; while (r - l > 1e-8) { double mid = (l + r)
阅读全文
摘要：#include <iostream> using namespace std; const int N = 1e5 + 10; int n, m; int q[N]; int main() { cin >> n >> m; for (int i = 0; i < n; ++ i) cin >> q
阅读全文
摘要：#include <iostream> using namespace std; const int N = 1e6 + 10; int n; int q[N], tmp[N]; long long cnt; void merge_sort(int q[], int l, int r) { if (
阅读全文
摘要：#include <iostream> using namespace std; const int N = 1e6 + 10; int n; int q[N], tmp[N]; void merge_sort(int q[], int l, int r) { if (l >= r) return
阅读全文
摘要：#include <iostream> using namespace std; const int N = 1e6 + 10; int n, k; int q[N]; int quick_sort(int l, int r, int k) { if (l == r) return q[l]; in
阅读全文
摘要：#include <iostream> using namespace std; const int N = 1e5 + 10; int n; int q[N]; void quick_sort(int q[], int l, int r) { if (l >= r) return ; int i
阅读全文
