wczffL_503/OI-Codes
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Frederick Chen 2024-10-03 19:17:29 +08:00
parent 543ccf0fe0
commit b64d31db74
39 changed files with 1548 additions and 0 deletions
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.DS_Store vendored
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61
FZOI/Tad 蛋糕.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
#define IMX LONG_LONG_MAX
#define IMN LONG_LONG_MIN
using namespace std;
/* toothless. #17 */
const int N = 1e7 + 10;
const int M = 2e3 + 5;
int n, m;
int a[1000000];
int qzh[1000000];
int st[1000000][25];
int lg[1000000];
void init()
{
lg[1] = 0;
for (int i = 2; i <= n; i++)
{
lg[i] = lg[i / 2] + 1;
}
for (int i = 1; i <= n; i++)
{
st[i][0] = qzh[i];
}
for (int k = 1; k <= lg[m]; k++)
{
for (int i = 1; i <= n; i++)
{
st[i][k] = max(st[i][k - 1], st[min(i + (1 << (k - 1)), n)][k - 1]);
}
}
}
int quary(int l, int r)
{
int k = lg[r - l + 1];
return max(st[l][k], st[r - (1 << k) + 1][k]);
}
signed main()
{
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
qzh[i] = qzh[i - 1] + a[i];
}
int maxl = -INT_MIN;
init();
for (int i = 1; i <= n; i++)
{
int r = min(i + m - 1, n);
int s = quary(i, r) - qzh[i - 1];
maxl = max(maxl, s);
}
cout << maxl;
return 0;
}

34
FZOI/The kth great number(对顶堆).cpp Normal file
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#include <bits/stdc++.h>
#define int long long
using namespace std;
signed main()
{
int n, k;
cin >> n >> k;
vector<int> nums;
for (int i = 0; i < n; ++i)
{
char op;
cin >> op;
if (op == 'I')
{
int num;
cin >> num;
nums.push_back(num);
}
else if (op == 'Q')
{
if (nums.size() >= k)
{
sort(nums.begin(), nums.end(), greater<int>());
cout << nums[k - 1] << endl;
}
}
}
return 0;
}

40
FZOI/grouping.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
#define IMX LONG_LONG_MAX
#define IMN LONG_LONG_MIN
using namespace std;
/* toothless. #17 */
const int N = 1e7 + 10;
const int M = 2e3 + 5;
int n, mn = IMX;
int a[N];
signed main()
{
cin >> n;
for (int i = 0; i < n; i ++ )
{
cin >> a[i];
}
sort(a, a + n);
int cnt = 1;
for (int i = 1; i < n; i ++ )
{
if (a[i] == a[i - 1] + 1)
{
cnt ++ ;
}
else
{
mn = min(mn, cnt);
cnt = 1;
}
}
mn = min(mn, cnt);
cout << mn << endl;
return 0;
}

24
FZOI/「NOIP2001」数的划分.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
using namespace std;
int n, k, cnt;
void dfs(int last, int sum, int cur)
{
if (cur == k)
{
if (sum == n)
cnt++;
return;
}
for (int i = last; sum + i * (k - cur) <= n; i++)
dfs(i, sum + i, cur + 1);
}
signed main()
{
cin >> n >> k ;
dfs(1, 0, 0);
cout << cnt ;
}

59
FZOI/「NOIP2004」合并果子.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
using namespace std;
const int N = 20005;
int n, heap[N], size;
void put(int t)
{
int p1, p2;
heap[ ++ size] = t;
p1 = size;
while(p1 > 1)
{
p2 = p1 / 2;
if(heap[p1] >= heap[p2]) return ;
swap(heap[p1], heap[p2]);
p1 = p2;
}
}
int out()
{
int p = 1, g, ans;
ans = heap[1];
heap[1] = heap[size -- ];
while(p * 2 <= size)
{
g = p * 2;
if(g < size && heap[g + 1] < heap[g]) g ++ ;
if(heap[p] <= heap[g]) return ans;
swap(heap[p], heap[g]);
p = g;
}
return ans;
}
signed main()
{
int ans = 0;
cin >> n ;
for(int i = 1; i <= n; ++ i)
{
int q;
cin >> q ;
put(q);
}
size = n;
while(size > 1)
{
int x = out(), y = out();
ans += (x + y);
put(x + y);
}
cout << ans ;
return 0;
}

29
FZOI/全排列问题.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
using namespace std;
int n, arr[1030], ans[10];
void dfs(int i, int s)
{
if (i > n)
{
for (int p = 1; p <= n; p++) printf("%5d", ans[p]);
cout << endl;
return;
}
for (int j = s; j > 0; j -= j & (-j))
{
int temp = j & (-j);
ans[i] = arr[temp];
dfs(i + 1, s - temp);
}
}
signed main()
{
cin >> n;
arr[1] = 1;
for (int i = 2; i <= n; i++)
arr[1 << (i - 1)] = i;
dfs(1, (1 << n) - 1);
return 0;
}

56
FZOI/分身术.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
#define fp(_a, _b, _c, _d) for (int _a = _b; _a <= _c; _a += _d)
#define fm(_a, _b, _c, _d) for (int _a = _b; _a <= _c; _a -= _d)
#define fin(_a, _b) for (int ss = 1; ss <= _a; ss++) cin >> _b[ss];
#define fout(_a, _b, _c) for (int ss = 1; ss <= _a; ss++) cout << _b[ss] << _c;
using namespace std;
/*
@author: BunDragon126
@link: https://www.setbun.com/
*/
const int N = 1e7 + 10;
const int M = 2e3 + 5;
int n, k;
vector<int> a(N), b(N);
deque<int> deque1, deque2;
signed main()
{
"toothless. #17";
cin >> n >> k;
for (int i = 0; i < n; ++ i) cin >> a[i];
for (int i = 0; i < n; ++ i) cin >> b[i];
int mn = n + 1;
for (int l = 0, r = 0; r < n; ++ r)
{
while (!deque1.empty() && a[deque1.back()] >= a[r])
deque1.pop_back();
deque1.push_back(r);
while (!deque2.empty() && b[deque2.back()] <= b[r])
deque2.pop_back();
deque2.push_back(r);
while (a[deque1.front()] <= k && b[deque2.front()] > a[deque1.front()])
{
mn = min(mn, r - l + 1);
if (deque1.front() == l)
deque1.pop_front();
if (deque2.front() == l)
deque2.pop_front();
++ l;
}
}
if (mn == n + 1)
{
cout << "So Sad!" << endl;
}
else
{
cout << mn << endl;
}
return 0;
}

56
FZOI/四色问题.cpp Normal file
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#include <iostream>
#include <cstring>
using namespace std;
const int MAXN = 8;
int n;
int arr[MAXN][MAXN];
int color[MAXN];
int ans;
bool f(int v, int c)
{
for (int i = 0; i < n; i++)
{
if (arr[v][i] && color[i] == c)
{
return false;
}
}
return true;
}
void dfs(int v)
{
if (v == n)
{
ans++;
return;
}
for (int c = 1; c <= 4; c++)
{
if (f(v, c))
{
color[v] = c;
dfs(v + 1);
color[v] = 0;
}
}
}
int main()
{
cin >> n;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
cin >> arr[i][j];
}
}
memset(color, 0, sizeof(color));
ans = 0;
dfs(0);
cout << ans << endl;
return 0;
}

57
FZOI/堆【模板题】.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
using namespace std;
const int N = 20005;
int n, heap[N], size;
void add(int t)
{
int p1, p2;
heap[ ++ size] = t;
p1 = size;
while(p1 > 1)
{
p2 = p1 / 2;
if(heap[p1] >= heap[p2]) return ;
swap(heap[p1], heap[p2]);
p1 = p2;
}
}
void out()
{
int p = 1, g, ans;
ans = heap[1];
heap[1] = heap[size -- ];
while(p * 2 <= size)
{
g = p * 2;
if(g < size && heap[g + 1] < heap[g]) g ++ ;
if(heap[p] <= heap[g]) return ;
swap(heap[p], heap[g]);
p = g;
}
}
signed main()
{
int ans = 0;
cin >> n ;
for(int i = 1; i <= n; ++ i)
{
int q;
cin >> q ;
if(q == 1)
{
int x;
cin >> x;
add(x);
}
else if(q == 2) cout << heap[1] << endl ;
else out();
}
return 0;
}

28
FZOI/奇怪的电梯.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
using namespace std;
int n, a, b, k[201], dis[201];
void dfs(int node, int step)
{
dis[node] = step;
int v = node - k[node];
if (1 <= v && step + 1 < dis[v])
dfs(v, step + 1);
v = node + k[node];
if (v <= n && step + 1 < dis[v])
dfs(v, step + 1);
return;
}
signed main()
{
memset(dis, 0x3f, sizeof(dis));
cin >> n >> a >> b;
for (int i = 1; i <= n; i++)
cin >> k[i];
dfs(a, 0);
if(dis[b] == 0x3f3f3f3f3f3f3f3f) cout << -1 << endl;
else cout << dis[b] << endl;
return 0;
}

36
FZOI/工作分配问题.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 20;
int n;
int arr[MAXN][MAXN];
int mn;
void f()
{
mn = INT_MAX;
for (int i = 0; i < n; i++)
{
int cnt = 0;
for (int j = 0; j < n; j++)
{
cnt += arr[j][i];
}
mn = min(mn, cnt);
}
}
signed main()
{
cin >> n;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
cin >> arr[i][j];
}
}
f();
cout << mn << endl;
return 0;
}

50
FZOI/括号匹配.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
using namespace std;
bool f(const string &s)
{
stack<char> st;
for (char c : s)
{
if (c == '(' || c == '[')
{
st.push(c);
}
else if (c == ')' || c == ']')
{
if (st.empty())
{
return false;
}
char top = st.top();
st.pop();
if ((c == ')' && top != '(') || (c == ']' && top != '['))
{
return false;
}
}
}
return st.empty();
}
signed main()
{
string s;
cin >> s;
if (f(s))
{
cout << "OK" << endl;
}
else
{
cout << "Wrong" << endl;
}
return 0;
}

30
FZOI/数位递增数.cpp Normal file
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#include <iostream>
#include <string>
using namespace std;
bool isIncreasingNumber(int n) {
string num = to_string(n);
for (int i = 0; i < num.size() - 1; i++) {
if (num[i] > num[i + 1]) {
return false;
}
}
return true;
}
int main() {
int n;
cin >> n;
int count = 0;
for (int i = 10; i <= n; i++) {
if (isIncreasingNumber(i)) {
count++;
}
}
cout << count << endl;
return 0;
}

63
FZOI/最少步数.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 100;
const vector<pair<int, int>> knight_moves = {{2, 1}, {2, -1}, {-2, 1}, {-2, -1}, {1, 2}, {1, -2}, {-1, 2}, {-1, -2}};
const vector<pair<int, int>> elephant_moves = {{2, 2}, {2, -2}, {-2, 2}, {-2, -2}};
int bfs(int start_x, int start_y)
{
vector<vector<bool>> vis(N + 1, vector<bool>(N + 1, false));
queue<tuple<int, int, int>> que;
que.push({start_x, start_y, 0});
vis[start_x][start_y] = true;
while (!que.empty())
{
auto [x, y, steps] = que.front();
que.pop();
if (x == 1 && y == 1)
{
return steps;
}
for (const auto &move : knight_moves)
{
int nx = x + move.first;
int ny = y + move.second;
if (nx >= 1 && nx <= N && ny >= 1 && ny <= N && !vis[nx][ny])
{
vis[nx][ny] = true;
que.push({nx, ny, steps + 1});
}
}
for (const auto &move : elephant_moves)
{
int nx = x + move.first;
int ny = y + move.second;
if (nx >= 1 && nx <= N && ny >= 1 && ny <= N && !vis[nx][ny])
{
vis[nx][ny] = true;
que.push({nx, ny, steps + 1});
}
}
}
return -1;
}
signed main()
{
int x, y, x1, y1;
cin >> x >> y >> x1 >> y1;
int steps_A = bfs(x, y);
int steps_B = bfs(x1, y1);
cout << steps_A << endl;
cout << steps_B << endl;
return 0;
}

80
FZOI/素数环.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
const int MAXN = 16;
int n;
int arr[MAXN];
bool vis[MAXN];
// 判断一个数是否为素数
bool check(int num)
{
if (num < 2)
{
return false;
}
for (int i = 2; i * i <= num; i++)
{
if (num % i == 0)
{
return false;
}
}
return true;
}
// 回溯
// cur表示当前位置
void back(int cur)
{
if (cur == n) // 检查最后一个数字和第一个数字之和是否为素数
{
if (check(arr[0] + arr[n - 1]))
{
for (int i = 0; i < n; i++)
{
cout << arr[i] << " ";
}
cout << endl;
}
return;
}
for (int i = 2; i <= n; i++)
{
if (!vis[i] && check(arr[cur - 1] + i))
{
vis[i] = true;
arr[cur] = i;
back(cur + 1);
vis[i] = false;
}
}
}
int main()
{
cin >> n;
memset(vis, false, sizeof(vis));
vis[1] = true;
arr[0] = 1;
back(1);
return 0;
}
/*
以下题解为后期由本人添加
题目分析：
题目要求将整数 1 到 n 组成一个环，使得相邻的两个整数之和均为素数。
我们需要找到所有满足条件的素数环，并按照要求输出。
由于 n 的值不超过 16，我们可以使用回溯法来尝试所有可能的组合。
代码解释：
check函数用于判断一个数是否为素数。
back函数是回溯函数，用于尝试放置数字到环中。
在back函数中，首先判断是否已经放置了 n 个数字。如果是，则检查最后一个数字和第一个数字之和是否为素数。如果是素数，则输出这个环。
然后，从数字 2 到 n 依次尝试将每个数字放在环的下一个位置。如果该数字没有被使用过，并且与前一个数字之和为素数，就将其放在环的下一个位置，并标记为已使用，然后继续递归放置下一个数字。
如果放置过程中出现相邻两个数字之和不是素数的情况，就回溯到上一个数字，重新选择下一个数字。
在main函数中，读取输入的 n，初始化标记数组和环的第一个数字，然后调用back函数从数字 1 开始回溯。
时间复杂度和空间复杂度：
时间复杂度：由于需要尝试所有可能的组合，时间复杂度为 O(n!)，其中 n 是输入的数字个数。
空间复杂度：主要是使用了标记数组和环数组，空间复杂度为 O(n)。
*/

43
FZOI/装载问题.cpp Normal file
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#include <iostream>
#include <cstring>
using namespace std;
const int MAXN = 45;
const int MAXC = 2000;
int n, c;
int arr[MAXN];
int dp[MAXN][MAXC];
int f()
{
for (int i = 0; i <= c; i++)
{
dp[0][i] = 0;
}
for (int i = 1; i <= n; i++)
{
for (int j = 0; j <= c; j++)
{
if (arr[i - 1] > j)
{
dp[i][j] = dp[i - 1][j];
}
else
{
dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - arr[i - 1]] + arr[i - 1]);
}
}
}
return dp[n][c];
}
int main()
{
cin >> n >> c;
for (int i = 0; i < n; i++)
{
cin >> arr[i];
}
cout << f() << endl;
return 0;
}

128
FZOI/非常可乐.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
struct Node
{
int s, n, m, step;
};
bool vis[101][101][101];
int bfs(int s, int n, int m)
{
queue<Node> que;
Node init = {s, 0, 0, 0};
que.push(init);
vis[s][0][0] = true;
while (!que.empty())
{
Node cur = que.front();
que.pop();
if (cur.s == cur.n && cur.s == cur.m)
{
return cur.step;
}
if (cur.s > 0 && cur.n < n)
{
int pour = min(cur.s, n - cur.n);
int newS = cur.s - pour;
int newN = cur.n + pour;
if (!vis[newS][newN][cur.m])
{
vis[newS][newN][cur.m] = true;
Node next = {newS, newN, cur.m, cur.step + 1};
que.push(next);
}
}
if (cur.s > 0 && cur.m < m)
{
int pour = min(cur.s, m - cur.m);
int newS = cur.s - pour;
int newM = cur.m + pour;
if (!vis[newS][cur.n][newM])
{
vis[newS][cur.n][newM] = true;
Node next = {newS, cur.n, newM, cur.step + 1};
que.push(next);
}
}
if (cur.n > 0 && cur.s < s)
{
int pour = min(cur.n, s - cur.s);
int newS = cur.s + pour;
int newN = cur.n - pour;
if (!vis[newS][newN][cur.m])
{
vis[newS][newN][cur.m] = true;
Node next = {newS, newN, cur.m, cur.step + 1};
que.push(next);
}
}
if (cur.n > 0 && cur.m < m)
{
int pour = min(cur.n, m - cur.m);
int newN = cur.n - pour;
int newM = cur.m + pour;
if (!vis[cur.s][newN][newM])
{
vis[cur.s][newN][newM] = true;
Node next = {cur.s, newN, newM, cur.step + 1};
que.push(next);
}
}
if (cur.m > 0 && cur.s < s)
{
int pour = min(cur.m, s - cur.s);
int newS = cur.s + pour;
int newM = cur.m - pour;
if (!vis[newS][cur.n][newM])
{
vis[newS][cur.n][newM] = true;
Node next = {newS, cur.n, newM, cur.step + 1};
que.push(next);
}
}
if (cur.m > 0 && cur.n < n)
{
int pour = min(cur.m, n - cur.n);
int newM = cur.m - pour;
int newN = cur.n + pour;
if (!vis[cur.s][newN][newM])
{
vis[cur.s][newN][newM] = true;
Node next = {cur.s, newN, newM, cur.step + 1};
que.push(next);
}
}
}
return -1;
}
int main()
{
int s, n, m;
while (cin >> s >> n >> m && !(s == 0 && n == 0 && m == 0))
{
memset(vis, false, sizeof(vis));
int res = bfs(s, n, m);
if (res == -1)
{
cout << "NO" << endl;
}
else
{
cout << res << endl;
}
}
return 0;
}

0
洛谷刷题/.DS_Store → Luogu/.DS_Store vendored
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0
Luogu/F_Chips_on_a_Line.cpp Normal file
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21
Luogu/Luogu_P_5715.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
using namespace std;
int arr[1005];
int n;
signed main()
{
for(int i = 1; i <= 3; i ++ )
{
cin >> arr[i];
}
sort(arr + 1, arr + 1 + 3);
for(int i = 1; i <= 3; i ++ )
{
cout << arr[i] << " ";
}
return 0;
}

119
Luogu/Luogu_P_9912.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
using namespace std;
const int N = 10001005;
int n, q, l, r, mid, x, sum, ma, ans[N], o, cnt, cnt1;
map<int, int> m;
// 结构体node用于存储线段树的节点信息
struct node {
int l, r, len, lc, rc; // l和r表示区间左右边界, len表示区间内岛屿的数量, lc和rc表示区间左右端点的类别
} a[N];
// 结构体deno用于存储高度信息
struct deno {
int x, id; // x表示高度, id表示高度对应的位置索引
} c[N];
// 结构体w用于存储查询信息
struct w {
int l, r, x, id; // l和r表示查询区间, x表示海平面上升高度, id表示查询的索引
} b[N];
// 更新节点信息函数
void gx(int p) {
a[p].len = a[p * 2].len + a[p * 2 + 1].len; // 合并左右子区间的岛屿数量
if (a[p * 2].rc == a[p * 2 + 1].lc) // 如果左右子区间的连接处高度相同，岛屿数量减少1
a[p].len--;
a[p].lc = a[p * 2].lc; // 更新左端点类别
a[p].rc = a[p * 2 + 1].rc; // 更新右端点类别
}
// 构建线段树函数
void build(int p, int l, int r) {
a[p].l = l;
a[p].r = r;
if (l == r) { // 叶子节点
a[p].len = a[p].lc = a[p].rc = 1; // 初始化为长度为1的岛屿
return;
}
int mid = (l + r) / 2;
build(p * 2, l, mid); // 递归构建左子树
build(p * 2 + 1, mid + 1, r); // 递归构建右子树
gx(p); // 更新当前节点信息
}
// 更新线段树函数
void change(int p, int l, int r) {
if (l <= a[p].l && a[p].r <= r) { // 完全覆盖
a[p].len = 0; // 将区间内的长度设置为0
a[p].lc = a[p].rc = ++cnt1; // 更新类别
return;
}
int mid = (a[p].l + a[p].r) / 2;
if (l <= mid) change(p * 2, l, r); // 递归更新左子树
if (mid < r) change(p * 2 + 1, l, r); // 递归更新右子树
gx(p); // 更新当前节点信息
}
// 查询线段树函数
int ask(int p, int l, int r) {
if (l <= a[p].l && a[p].r <= r) // 完全覆盖
return a[p].len;
int mid = (a[p].l + a[p].r) / 2;
int ans1 = 0;
if (l <= mid && mid < r) { // 跨区间查询
ans1 = ask(p * 2, l, r) + ask(p * 2 + 1, l, r); // 查询左右子树
if (a[p * 2].rc == a[p * 2 + 1].lc) // 如果左右子树的连接处高度相同，岛屿数量减少1
ans1--;
} else if (l <= mid) { // 查询左子区间
ans1 = ask(p * 2, l, r);
} else if (mid < r) { // 查询右子区间
ans1 = ask(p * 2 + 1, l, r);
}
return ans1;
}
// 比较函数，用于排序查询
bool cmp(w x, w y) {
return x.x < y.x;
}
// 比较函数，用于排序高度
bool cmp1(deno x, deno y) {
return x.x < y.x;
}
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> q;
cnt1 = n + 1;
for (int i = 1; i <= n; i++) {
cin >> c[i].x;
c[i].id = i;
}
sort(c + 1, c + 1 + n, cmp1); // 按高度排序
for (int i = 1; i <= q; i++) {
cin >> b[i].l >> b[i].r >> b[i].x;
b[i].id = i;
}
sort(b + 1, b + 1 + q, cmp); // 按海平面上升高度排序
cnt = 1;
build(1, 1, n); // 构建线段树
for (int i = 1; i <= q; i++) {
while (cnt <= n && c[cnt].x <= b[i].x) {
change(1, c[cnt].id, c[cnt].id); // 更新线段树
cnt++;
}
ans[b[i].id] = ask(1, b[i].l, b[i].r); // 查询岛屿数量
}
for (int i = 1; i <= q; i++) {
cout << ans[i] << endl; // 输出结果
}
return 0;
}

53
Luogu/Luogu_U_458232.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
#define fp(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a += _d)
#define fm(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a -= _d)
#define fin(_a, _b) for(int ss = 1; ss <= _a; ss ++ ) cin >> _b[ss];
#define fout(_a, _b, _c) for(int ss = 1; ss <= _a; ss ++ ) cout << _b[ss] << _c ;
using namespace std;
/*
2024/08/03
Code-From-BunDragon126-At-VisualStudioCode
AlignPixel*GiveSupport
https://frederication.work/
*/
const int N = 1e7 + 10;
const int M = 2e3 + 5;
int t, n, m, k, a[N];
signed main()
{
"toothless. #17";
cin >> t;
fp(ss, 1, t, 1)
{
cin >> n >> m ;
if(m % 2 == 0) // 如果 m 是偶数且 n 为 0 ，输出 Yes。否则输出 No。
{
if(n == 0)
{
cout << "Yes" << endl;
}
else
{
cout << "No" << endl;
}
}
else // 如果 m 是奇数且 n 为正整数，输出 Yes 。否则输出 No。
{
if(n > 0)
{
cout << "Yes" << endl;
}
else
{
cout << "No" << endl;
}
}
}
return 0;
}

58
Luogu/Luogu_U_458233.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
#define fp(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a += _d)
#define fm(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a -= _d)
#define fin(_a, _b) for(int ss = 1; ss <= _a; ss ++ ) cin >> _b[ss];
#define fout(_a, _b, _c) for(int ss = 1; ss <= _a; ss ++ ) cout << _b[ss] << _c ;
using namespace std;
/*
<DATE>
Code-From-BunDragon126-At-VisualStudioCode
AlignPixel*GetSupport
https://frederication.work/
*/
const int N = 1e7 + 10;
const int M = 2e3 + 5;
signed main()
{
"toothless. #17";
int n, m;
cin >> n >> m;
string a, b;
cin >> a >> b;
char current_char = 'a';
fp(i, 0, m - 1, 1)
{
bool found = false;
fp(j, 0, n - 1, 1)
{
if (a[j] == '#')
{
a[j] = current_char;
found = true;
break;
}
}
if (!found) {
fp(j, 0, n - 1, 1)
{
if (b[j] == '#')
{
b[j] = current_char;
break;
}
}
}
current_char ++ ;
if(current_char > 'z')
{
current_char = 'a';
}
}
cout << a << endl;
return 0;
}

64
Luogu/Luogu_U_458234.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
#define fp(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a += _d)
#define fm(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a -= _d)
#define fin(_a, _b) for(int ss = 1; ss <= _a; ss ++ ) cin >> _b[ss];
#define fout(_a, _b, _c) for(int ss = 1; ss <= _a; ss ++ ) cout << _b[ss] << _c ;
using namespace std;
/*
2024/08/03
Code-From-BunDragon126-At-VisualStudioCode
AlignPixel*GetSupport
https://frederication.work/
*/
const int N = 1e7 + 10;
const int M = 2e3 + 5;
int calculateUniqueBloomTimes(vector<int>& times, int m, int offset = 0) {
unordered_map<int, int> bloomCount;
for (int t : times) {
for (int i = 0; i < m; ++i) {
bloomCount[t + offset + i]++;
}
}
int uniqueBloomTimes = 0;
for (auto& entry : bloomCount) {
if (entry.second == 1) {
uniqueBloomTimes++;
}
}
return uniqueBloomTimes;
}
signed main()
{
"toothless. #17";
int n, m;
cin >> n >> m;
vector<int> t(n);
for (int i = 0; i < n; ++i) {
cin >> t[i];
}
int maxUniqueBloomTimes = calculateUniqueBloomTimes(t, m);
for (int i = 0; i < n; ++i) {
int originalTime = t[i];
for (int newTime = 1; newTime <= *max_element(t.begin(), t.end()) + m; ++newTime) {
if (newTime != originalTime) {
t[i] = newTime;
maxUniqueBloomTimes = max(maxUniqueBloomTimes, calculateUniqueBloomTimes(t, m));
}
}
t[i] = originalTime;
}
cout << maxUniqueBloomTimes << endl;
return 0;
}

0
洛谷刷题/P1005 [NOIP2007 提高组] 矩阵取数游戏.cpp → Luogu/P1005 [NOIP2007 提高组] 矩阵取数游戏.cpp
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26
Luogu/P_1001_A_B_Problem.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
#define fp(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a += _d)
#define fm(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a -= _d)
#define fin(_a, _b) for(int ss = 1; ss <= _a; ss ++ ) cin >> _b[ss];
#define fout(_a, _b, _c) for(int ss = 1; ss <= _a; ss ++ ) cout << _b[ss] << _c ;
using namespace std;
/*
@author: BunDragon126
@link: https://www.setbun.com/
*/
const int N = 1e7 + 10;
const int M = 2e3 + 5;
int a, b;
signed main()
{
"toothless. #17";
cin >> a >> b ;
cout << a + b ;
return 0;
}

0
Luogu/P_1002_NOIP_2002_普及组_过河卒.cpp Normal file
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22
Luogu/P_1085_NOIP_2004_普及组_不高兴的津津.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
#define fp(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a += _d)
#define fm(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a -= _d)
#define fin(_a, _b) for(int ss = 1; ss <= _a; ss ++ ) cin >> _b[ss];
#define fout(_a, _b, _c) for(int ss = 1; ss <= _a; ss ++ ) cout << _b[ss] << _c ;
using namespace std;
signed main()
{
int a, b, s, max = 0, i, day = 0;
fp(i, 1, 8, 1)
{
cin >> a >> b;
s = a + b;
if ((s > max) && (s > 8))
max = s, day = i;
}
cout << day;
return 0;
}

46
Luogu/P_1089_NOIP_2004_提高组_津津的储蓄计划.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
#define fp(_a, _b, _c, _d) for (int _a = _b; _a <= _c; _a += _d)
#define fm(_a, _b, _c, _d) for (int _a = _b; _a <= _c; _a -= _d)
#define fin(_a, _b) for (int ss = 1; ss <= _a; ss++) cin >> _b[ss];
#define fout(_a, _b, _c) for (int ss = 1; ss <= _a; ss++) cout << _b[ss] << _c;
using namespace std;
/*
@author: BunDragon126
@link: https://www.setbun.com/
*/
const int N = 1e7 + 10;
const int M = 2e3 + 5;
int arr[N];
signed main()
{
"toothless. #17";
for (int i = 0; i < 12; i++)
{
cin >> arr[i];
}
int money = 0;
int l = 0;
for (int i = 0; i < 12; i++)
{
money += 300;
if (money < arr[i])
{
cout << -(i + 1) << endl;
return 0;
}
money -= arr[i];
int c = (money / 100) * 100;
l += c;
money -= c;
}
cout << money + l * 1.2 << endl;
return 0;
}

32
Luogu/P_1150_Peter_的烟.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
#define fp(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a += _d)
#define fm(_a, _b, _c, _d) for(int _a = _b; _a <= _c; _a -= _d)
#define fin(_a, _b) for(int ss = 1; ss <= _a; ss ++ ) cin >> _b[ss];
#define fout(_a, _b, _c) for(int ss = 1; ss <= _a; ss ++ ) cout << _b[ss] << _c ;
using namespace std;
/*
@author: BunDragon126
@link: https://www.setbun.com/
*/
const int N = 1e7 + 10;
const int M = 2e3 + 5;
int n, k;
signed main()
{
"toothless. #17";
cin >> n >> k ;
int cnt = 0;
fp(i, 1, n, 1)
{
cnt ++ ;
if(cnt % k == 0) n ++ ;
}
cout << cnt ;
return 0;
}

0
Luogu/P_1739_表达式括号匹配.cpp Normal file
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0
Luogu/U_458233_FLA_I_歌静河.cpp Normal file
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54
Luogu/[COCI 2023:2024 #2] Zatopljenje.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
const int N = 10005;
vector<int> h(N);
int f(const vector<int> &h, int l, int r, int x)
{
int cnt = 0;
int start = l - 1;
bool flag = false;
for (int i = start; i < r; ++i)
{
if (h[i] > x)
{
if (!flag)
{
cnt++;
flag = true;
}
}
else
{
flag = false;
}
}
return cnt;
}
signed main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, q;
cin >> n >> q;
for (int i = 0; i < n; ++i)
{
cin >> h[i];
}
for (int i = 0; i < q; ++i)
{
int l, r, x;
cin >> l >> r >> x;
cout << f(h, l, r, x) << endl ;
}
return 0;
}

10
Luogu/小鱼的航程.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
int main(){
long long x, n;
cin >> x >> n;
long long week;
week = (x + n) / 7 * 2;
cout << (n - week) * 250;
return 0;
}

36
Luogu/数字三角形2.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
using namespace std;
const int N = 2e3 + 10;
int mp[N][N], d[N][N];
signed main()
{
int n;
cin >> n ;
for(int i = 1; i <= n; ++ i)
{
for(int j = 1; j <= i; ++ j)
{
cin >> mp[i][j] ;
}
}
for(int i = 1; i <= n; ++ i)
{
for(int j = 1; j <= i; ++ j)
{
d[i][j] = max(d[i - 1][j - 1] + mp[i][j] % 100, d[i - 1][j] + mp[i][j] % 100);
}
}
int ans = 0;
for(int i = 1; i <= n; ++ i)
{
// cout << d[n][i] << " " ;
ans = max(ans, d[n][i] % 100);
}
cout << ans ;
return 0;
}

85
Luogu/素数环【洛谷版UVa】.cpp Normal file
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#include <bits/stdc++.h>
using namespace std;
const int MAXN = 16;
int n;
int arr[MAXN];
bool vis[MAXN];
// 判断一个数是否为素数
bool check(int num)
{
if (num < 2)
{
return false;
}
for (int i = 2; i * i <= num; i++)
{
if (num % i == 0)
{
return false;
}
}
return true;
}
// 回溯
// cur表示当前位置
void back(int cur)
{
if (cur == n) // 检查最后一个数字和第一个数字之和是否为素数
{
if (check(arr[0] + arr[n - 1]))
{
for (int i = 0; i < n; i++)
{
cout << arr[i] << " ";
}
cout << endl;
}
return;
}
for (int i = 2; i <= n; i++)
{
if (!vis[i] && check(arr[cur - 1] + i))
{
vis[i] = true;
arr[cur] = i;
back(cur + 1);
vis[i] = false;
}
}
}
int main()
{
int idx = 1;
while(n != EOF)
{
cin >> n;
cout << "Case " << idx++ << ":" << endl;
memset(vis, false, sizeof(vis));
vis[1] = true;
arr[0] = 1;
back(1);
}
return 0;
}
/*
以下题解为后期由本人添加
题目分析：
题目要求将整数 1 到 n 组成一个环，使得相邻的两个整数之和均为素数。
我们需要找到所有满足条件的素数环，并按照要求输出。
由于 n 的值不超过 16，我们可以使用回溯法来尝试所有可能的组合。
代码解释：
check函数用于判断一个数是否为素数。
back函数是回溯函数，用于尝试放置数字到环中。
在back函数中，首先判断是否已经放置了 n 个数字。如果是，则检查最后一个数字和第一个数字之和是否为素数。如果是素数，则输出这个环。
然后，从数字 2 到 n 依次尝试将每个数字放在环的下一个位置。如果该数字没有被使用过，并且与前一个数字之和为素数，就将其放在环的下一个位置，并标记为已使用，然后继续递归放置下一个数字。
如果放置过程中出现相邻两个数字之和不是素数的情况，就回溯到上一个数字，重新选择下一个数字。
在main函数中，读取输入的 n，初始化标记数组和环的第一个数字，然后调用back函数从数字 1 开始回溯。
时间复杂度和空间复杂度：
时间复杂度：由于需要尝试所有可能的组合，时间复杂度为 O(n!)，其中 n 是输入的数字个数。
空间复杂度：主要是使用了标记数组和环数组，空间复杂度为 O(n)。
*/

48
Luogu/螺旋矩阵.cpp Normal file
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#include <bits/stdc++.h>
#define int long long
#define endl "\n"
using namespace std;
signed main()
{
int n, x, y;
cin >> n >> x >> y ;
int rx = 1, num = y; // rx为模拟到的行数，num为目前加到的数，也是最终输出的结果
int up = 1, down = n, zuo = 1, you = n; // x、y在下一次绕圈时上下左右的边界值（可以理解为将要被蛇形填数的一圈的边界）
bool fx = true; // 方向（true往下，false往上）
while (rx != x) // 一直模拟到刚好到达
{
if (fx) // 特判
{
if (you == y) // 如果是往下走直线
{
num += x - rx; // 加上最后的数
break;
}
}
else
{
if (zuo == y) // 如果是向上走直线
{
num += rx - x;
break;
}
}
if (fx) // 普通情况，如果向下走
{
num += (you - y + 1) * 2 + down - rx - 2; // 绕一圈
rx = down; // 更新所在的行
you -- ; // 右边界往回退一格
up ++ ; // 上边界往回退一格
}
else // 往上走
{
num += (y - zuo + 1) * 2 + rx - up - 2; // 绕一圈
rx = up; // 更新行
zuo ++ ; // 左边界向内
down -- ; // 下边界向内
}
fx = !fx; // 方向调转
}
cout << num ;
return 0;
}

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XSM OJ 重庆小码王集团OJ/.DS_Store vendored
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