A.质数
题面
思路
考虑到输入数据量较大，选择线性欧拉筛预处理
示例代码
#include<bits/stdc++.h>using namespace std;#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'const int MOD = 1e9 + 7;
const int N = 1e5 + 2;
const int inf = 1e9;vector<bool> is_prime(N, true);
vector<int> primes;void siebe() {is_prime[0] = is_prime[1] = false;fer(i, 2, N) {if(is_prime[i]) primes.push_back(i);for(int j = 0; j < primes.size() && i * primes[j] < N; ++j) {is_prime[i * primes[j]] = false;if(i % primes[j] == 0) break;}}
}signed main() {ios::sync_with_stdio(false); cin.tie(nullptr);siebe();int t;cin >> t;while(t--){int n;cin >> n;cout << (is_prime[n] ? "yes" : "no");cout << '\n';}return 0;
}
## [B.分治法求解全排列问题](https://buctoj.com/problem.php?cid=3923&pid=1)
### 题面
### 思路
分治法求全排列问题，注意输出的答案不兼容其他做法（如果是按字典序输出的做法不可行）
### 示例代码
```cpp
#include<bits/stdc++.h>using namespace std;#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;void perm(int n, int l, int r, vector<int>& arr) {if(l == r) {fer(i, 0, n) cout << arr[i] << " ";cout << '\n';return;}for(int i = l; i <= r; ++i) {swap(arr[l], arr[i]);perm(n, l + 1, r, arr);swap(arr[l], arr[i]);}
}signed main() {ios::sync_with_stdio(false); cin.tie(nullptr);int n;while(cin >> n) {vector<int> arr(n);fer(i, 0, n) arr[i] = i + 1;perm(n, 0, n - 1, arr);}return 0;
}
C.数的计数
题面
思路
递归处理，在处理过程中同时进行记忆化。
示例代码
#include<bits/stdc++.h>using namespace std;#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'const int MOD = 1e9 + 7;
const int N = 1e3 + 2;
const int inf = 1e9;vector<int> memo(N, -1);int generateNumbers(int n) {if (n == 0) return 0;if (memo[n] != -1) return memo[n];int cnt = 1;for (int i = 1; i <= n / 2; ++i) cnt += generateNumbers(i);memo[n] = cnt;return cnt;
}signed main() {ios::sync_with_stdio(false); cin.tie(nullptr);int n;cin >> n;cout << generateNumbers(n) << '\n';return 0;
}
D.循环日程表问题，《算法竞赛入门经典》P230，递归与分治，刘丽萍，CCF三级
题面
示例代码
#include <bits/stdc++.h>
using namespace std;
#define int long longlong long mergeAndCount(vector<int>& arr, int left, int mid, int right) {int n1 = mid - left + 1;int n2 = right - mid;vector<int> L(n1), R(n2);for (int i = 0; i < n1; ++i)L[i] = arr[left + i];for (int j = 0; j < n2; ++j)R[j] = arr[mid + 1 + j];int i = 0, j = 0, k = left, swaps = 0;while (i < n1 && j < n2) {if (L[i] <= R[j]) {arr[k++] = L[i++];} else {arr[k++] = R[j++];swaps += (mid + 1) - (left + i);}}while (i < n1) {arr[k++] = L[i++];}while (j < n2) {arr[k++] = R[j++];}return swaps;
}int mergeSortAndCount(vector<int>& arr, int left, int right) {long long count = 0;if (left < right) {int mid = left + (right - left) / 2;count += mergeSortAndCount(arr, left, mid);count += mergeSortAndCount(arr, mid + 1, right);count += mergeAndCount(arr, left, mid, right);}return count;
}signed main() {ios::sync_with_stdio(false); cin.tie(nullptr);int n;cin >> n;vector<int> arr(n);for(int i = 0; i < n; ++i) {cin >> arr[i];}int result = mergeSortAndCount(arr, 0, n - 1);cout << result << endl;return 0;
}
E.求逆序对(deseq)
题面
示例代码
#include <bits/stdc++.h>
using namespace std;
#define int long longlong long mergeAndCount(vector<int>& arr, int left, int mid, int right) {int n1 = mid - left + 1;int n2 = right - mid;vector<int> L(n1), R(n2);for (int i = 0; i < n1; ++i)L[i] = arr[left + i];for (int j = 0; j < n2; ++j)R[j] = arr[mid + 1 + j];int i = 0, j = 0, k = left, swaps = 0;while (i < n1 && j < n2) {if (L[i] <= R[j]) {arr[k++] = L[i++];} else {arr[k++] = R[j++];swaps += (mid + 1) - (left + i);}}while (i < n1) {arr[k++] = L[i++];}while (j < n2) {arr[k++] = R[j++];}return swaps;
}int mergeSortAndCount(vector<int>& arr, int left, int right) {long long count = 0;if (left < right) {int mid = left + (right - left) / 2;count += mergeSortAndCount(arr, left, mid);count += mergeSortAndCount(arr, mid + 1, right);count += mergeAndCount(arr, left, mid, right);}return count;
}signed main() {ios::sync_with_stdio(false); cin.tie(nullptr);int n;cin >> n;vector<int> arr(n);for(int i = 0; i < n; ++i) {cin >> arr[i];}int result = mergeSortAndCount(arr, 0, n - 1);cout << result << endl;return 0;
}
