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xingyou
2025-03-14 22:39:32 +08:00
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#include<bits/stdc++.h>
using namespace std;
const int N = 1e2 + 10;
int a[N] = { 0,5,6,8,6,9,1,6,1,2,4,9,1,9,8,2,3,6,4,7,7,5,9,5,0,3,8,7,5,8,1,5,8,6,1,8,3,0,3,7,9,2,
7,0,5,8,8,5,7,0,9,9,1,9,4,4,6,8,6,3,3,8,5,1,6,3,4,6,7,0,7,8,2,7,6,8,9,5,6,5,6,1,4,0,1,
0,0,9,4,8,0,9,1,2,8,5,0,2,5,3,3 }; // 示例数组
int day[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int main(){
int ans = 0;
for(int i = 1; i <=12; i++){
for(int j = 1; j <= day[i]; j++){
int date[] = {0, 2, 0, 2, 3, i/10, i%10, j/10, j%10 };
int id = 1;
for(int i = 1; i<=100; i++){
if(a[i] == date[id]){
id++;
if(id>8){
ans++;
break;
}
}
}
}
}
cout << ans << endl;
return 0;
}

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#include<bits/stdc++.h>
using namespace std;
int main(){
double n = 23333333;
double res = 11625907.5798;
for(int a = 0; a <= n; a++){ // a为0出现的次数
int b = n - a; // b为1出现的次数
if(a>=b) continue;
double ans = (-1)*a*(a/n)*log2(a/n) + (-1)*b*(b/n)*log2(b/n);
if(fabs(ans - res) < 1e-4) cout << a << endl;
}
return 0;
}

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14lanqiao/test3-1.cpp Normal file
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#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e4 + 10;
int a[N], b[N];
//枚举法
signed main(){
int n; cin >> n;
int mxv = LLONG_MIN, miv = LLONG_MAX;
for(int i = 1; i <= n; i++){
cin >> a[i] >> b[i];
if(b[i] != 1) mxv = max(mxv, a[i]/(b[i] - 1));
miv = min(miv, a[i]/(b[i] + 1));
}
for(int v = miv; v <= mxv; v++){
bool flag = true;
for(int i = 1; i <= n; i++){
if(a[i] / v != b[i]){
flag = false;
break;
}
}
if(flag){
cout << v << " ";
break;
}
}
for(int v = mxv; v >= miv; v--){
bool flag = true;
for(int i = 1; i <= n; i++){
if(a[i] / v != b[i]){
flag = false;
break;
}
}
if(flag){
cout << v << endl;
break;
}
}
return 0;
}
/* test samples
3
75 3
53 2
59 2
*/

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14lanqiao/test3-2.cpp Normal file
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#include<bits/stdc++.h>
using namespace std;
int main(){
int n; cin >> n;
int mxv = INT_MIN; int miv = INT_MAX;
for(int i = 1; i <= n; i++){
int a, b; cin >> a >> b;
miv = min(miv, a/b);
mxv = max(mxv, a/(b+1)+1);
}
cout << mxv << " " << miv << endl;
return 0;
}
/*
数学优化
通过floor(A/V) = B，可以推出 A/V =［B,B+1）
进而得到B <= A/V < B+1，三个表达式同时被A除
得到关于V的不等式即 A/B >= V > A/(B+1)
*/
/* test samples
3
75 3
53 2
59 2
*/

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#include<bits/stdc++.h>
using namespace std;
const int N = 1e2 + 10;
struct node{
int t, d, l; // 分别代表可以降落的时刻，盘旋时间，降落花费时间
}a[N];
bool vis[N];
int t, n;
bool flag = false;
// lasttime: 之前完成降落的所有飞机中的最后一架飞机降落的时间
void dfs(int dep, int lasttime){
if(dep == n + 1){
flag = true;
return;
}
for(int i = 1; i <= n; i++){
if(!vis[i] && a[i].t + a[i].d >= lasttime){
vis[i] = 1;
dfs(dep + 1, max(lasttime, a[i].t) + a[i].l);
// 遍历同级的另一个飞机的下一个深度的飞机前，需要将该飞机设置为未安排状态
vis[i] = 0;
}
}
}
int main(){
cin >> t;
while(t--){
cin >> n;
for(int i = 1; i<=n; i++){
cin >> a[i].t >> a[i].d >> a[i].l;
}
flag = false;
memset(vis, 0, sizeof(vis));
dfs(1, 0);
if(flag) cout << "YES" << endl;
else cout << "NO" << endl;
}
return 0;
}
/* test samples
2
3
0 100 10
10 10 10
0 2 20
3
0 10 20
10 10 20
20 10 20
*/

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14lanqiao/test5-1.cpp Normal file
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#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int ft[N], bk[N], dp[N];
/*
ft[i]第i个元素的首数字、bk［i］第i个元素尾数字
类LIS问题的DP解法-时间复杂度0(n^2)
状态 dp[i] 以第主个元素作为结尾的最长接龙数列的长度
状态转移方程 dp[i] = max(dp[i], dp[j]+1);
*/
int main(){
int n; cin >> n;
string s;
for(int i = 1; i<=n; i++){
cin >> s;
ft[i] = s.front() - '0', bk[i] = s.back() - '0', dp[i] = 1;
}
int mx = 1;
for(int i = 2; i <= n; i++){
for(int j = 1; j < i; j++){
if(bk[j] == ft[i]){
dp[i] = max(dp[i], dp[j] + 1);
}
mx = max(mx, dp[i]);
}
}
cout << n - mx << endl;
return 0;
}
/* test samples 易超时
5
11 121 22 12 2023
*/

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14lanqiao/test5-2.cpp Normal file
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#include<bits/stdc++.h>
using namespace std;
int dp[10];
/*
优化dp方案
状态 dp[i] 以数字i结尾的最长接龙序列的长度
状态转移方程 dp[bk]=max(dp[ft]+1, dp[bk])
*/
int main(){
int n; cin >> n;
string s;
int mx = 1;
for(int i = 0; i < n; i++){
cin >> s;
int ft = s.front() - '0', bk = s[s.size() - 1] - '0';
dp[bk] = max(dp[ft] + 1, dp[bk]);
mx = max(mx, dp[bk]);
}
cout << n - mx << endl;
return 0;
}
/* test samples
5
11 121 22 12 2023
*/

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15lanqiao/test1-1.cpp Normal file
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#include<bits/stdc++.h>
using namespace std;
int main() {
int tot = (50*49) / 2, cnt = (7*6) / 2;
cout << tot - cnt << endl;
return 0;
}
/*
总共有50人参与会议，如果按照常规情况，即每个人与除了自己以外的所有人各握手一次
握手总数tot=49+48+...+1=(50*49)/2
现在有7个人没有互相握手，这7这个人本来应该握手的次数 cnt=6+5+...+1=(7*6)/2
实际握手次数=总的握手次数tot-7人的情况cnt (tot - cnt)
*/

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#include<bits/stdc++.h>
using namespace std;
int main() {
int ans = 0;
for(int i = 1; i <=50; i++){
for(int j = i + 1; j<=50; j++){
if(!( (i >= 1 && i<= 7) && (j >= 1 && j <= 7) )) ans++;
// i和j不能同时满足都在[1, 7]范围内
}
}
cout << ans << endl;
return 0;
}
/* 枚举法
*/

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#include<bits/stdc++.h>
using namespace std;
int main() {
int x = 343720, y = 233333, dx = 15, dy = 17;
int p = y*dx, q = x*dy;
int g = gcd(p, q);
p /= g, q /= g;
int t = 2*p*x / dx;
double ans = t * (sqrt(15*15 + 17*17));
printf("%.2lf\n", ans);
return 0;
}
/*
思路:针对前进的方向进行分解，分为x方向的运动和y方向的运动假设x方向走了p个来回，y反向走了q个来回，
经过了时间t，小球第一次回到原点则t*dx=2px,t*dy=2qy，结合1式/2式，得p/q=y/x*dx/dy
利用gcd对分式p,q进行约分，进而得到约分后的p,q
则时间 t = 2px/dx, 总路程 = t * sqrt(15*15 + 17*17)
*/

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#include<bits/stdc++.h>
using namespace std;
bool check(int n){
int b = 1; //数位
while(n){
if(b % 2){ // 奇数位
if(!(n % 10 % 2)) return false;
}else{ // 偶数位
if(n % 10 % 2) return false;
}
n /= 10;
b++;
}
return true;
}
int main() {
int n; cin >>n;
int ans = 0;
for(int i = 1; i <= n; i++){
if(i % 10 % 2 == 0) continue;
if(check(i)) ans++;
}
cout << ans << endl;
return 0;
}
/*
一个整数如果按从低位到高位的顺序，奇数位为奇数，偶数位为偶数，则该数为好数
24 -> 7, 2024 -> 150
*/

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#include<bits/stdc++.h>
using namespace std;
const int N = 2e3 + 10;
string s; int a[N];
int main() {
int n; cin >> n >> s;
reverse(s.begin(), s.end());
int pos = s.find("."); // 小数点位置
s.erase(pos, 1); // 删除小数点, 方便后续计算
int len = s.size();
for(int i = 0; i < len; i++) a[i+1] = s[i] - '0';
// a[1] = 4，a[2] = 1，a[3] = 3
//高精度*低精度模版
for(int i = 1; i<=n; i++){
// 顺序扫描每一位, 均乘2
for(int j = 1; j <= len; j++) a[j] *= 2;
for(int j = 1; j <= len; j++){
if(a[j] > 10){
a[j+1]++;
a[j] %= 10;
if(j == len) len++;
}
}
}
// a[1] = 6，a[2] = 5，a[3] = 2，a[4] = 1 a[pos] = 5
if(a[pos] >= 5) a[pos+1]++;
for(int i = len; i >= pos + 1; i--) cout << a[i];
cout << endl;
return 0;
}
/*
2 3.14
*/

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#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int a[N];
vector<int> fac[N], s[N];
int main() {
for(int i = 1; i <= 1e5; i++){
for(int j = i; j <= 1e5; j += i){
fac[j].push_back(i);
}
}
int n; cin>>n;
for(int i = 1; i<=n; i++) cin >> a[i];
// 保证字典序最小
sort(a+1, a+1+n);
for(int i = 1; i <=n; i++){
for(int j = 0; j < fac[a[i]].size(); j++){
s[fac[a[i]][j]].push_back(a[i]);
}
}
for(int i = 1e5; i>=0; i--){
if(s[i].size() >= 3){
cout << s[i][0] << " " << s[i][1] << " " << s[i][2] << endl;
break;
}
}
return 0;
}
/* test samples
5
1 2 3 4 9
*/
/*
排序后的宝石为 [1, 2, 3, 4, 9]。
因数数组 fac 和宝石数组 s 的关系如下：
s[1] 包含所有宝石 [1, 2, 3, 4, 9]。
s[2] 包含 [2, 4]。
s[3] 包含 [3, 9]。
s[4] 包含 [4]。
s[9] 包含 [9]。
最终，s[1] 中有5个宝石，满足条件。
输出前三个宝石：1 2 3。
*/

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#include<bits/stdc++.h>
using namespace std;
const int N = 20;
int a[N][N]; // 迷宫数组
int vis[N][N]; // 标记数组
int n, k;
// 方向数组 0 1 2 3 4 5 6 7
int dx[] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dy[] = { 0, 1, 1, 1, 0, -1, -1, -1};
string res; // 结果字符串
void dfs(int x, int y, int pre, string s, int dep){
if(x == n && y == n && dep == n*n){
if(res.empty()) res = s; // 字典需最小的
return;
}
for(int i = 0; i < 8; i++){
int bx = x + dx[i], by = y + dy[i];
if(bx < 1 || bx > n || by < 1 || by > n) continue;
if(vis[bx][by]) continue;
// 防止交叉搜索
if(i == 1 && vis[x-1][y] && vis[x][y+1]) continue;
else if(i == 3 && vis[x+1][y] && vis[x][y+1]) continue;
else if(i == 5 && vis[x+1][y] && vis[x][y-1]) continue;
else if(i == 7 && vis[x-1][y] && vis[x][y-1]) continue;
// 保证路径数值为0 1 ... k-1
if((a[bx][by] < k && a[bx][by] == pre + 1) || (pre + 1 == k && a[bx][by] == 0)){
vis[bx][by] = 1;
dfs(bx, by, a[bx][by], s + to_string(i), dep + 1);
// 最优性剪枝
if(!res.empty()) return;
vis[bx][by] = 0; //回溯
}
}
}
int main() {
cin >> n >> k;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
cin >> a[i][j];
}
}
string emp;
vis[1][1] = 1;
dfs(1, 1, 0, emp, 1);
if(res.empty()) cout << -1 << endl;
else cout << res << endl;
return 0;
}
/* test samples
3 3
0 2 0
1 1 1
2 0 2
*/

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#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int main(){
priority_queue<int> pq; // 大根堆
int n, p ,q; cin >> n >> p >> q;
for(int i = 1; i <= n; i++){
int x; cin >> x;
pq.push(x);
}
while(p || q){
int cur = pq.top(); // 最大值
pq.pop();
if(p){ cur = sqrt(cur); p--; }
else if(q) { cur = cur / 2; q--; }
pq.push(cur);
}
int s = 0; // 总的消耗
while(!pq.empty()) { s += pq.top(); pq.pop(); }
cout << s << endl;
return 0;
}
/* test samples
4 1 1
4 5 6 49
*/

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#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e3+10;
int a[N], s[N]; // s为前缀和数组
multiset<int> ms;
signed main(){
int n; cin>>n;
for(int i = 1; i<=n; i++){
cin >> a[i];
s[i] = s[i-1] + a[i];
}
// 使用set数组去维护所有的区间和
for(int i = 1; i<=n; i++){
for(int j = 1; j<=n; j++){
ms.insert(s[j] - s[i-1]);
}
}
int ans = LLONG_MAX;
// 时间复杂度为 O(n^log2n)
for(int i = 1; i<=n; i++){
for(int j = 1; j<i; j++){
// 枚举以i结尾的区间
int sum = s[i] - s[j-1];
// 找到该区间和sum相似的区间和s1 and s2
auto it = ms.lower_bound(sum);
if(it!=ms.end()){ ans = min(ans, abs(*it - sum)); } // 找到了
if(it!=ms.begin()){
it--;
ans = min(ans, abs(*it - sum));
}
}
// 删除以i开头且以j结尾的区间，防止后续查询区间的时候出现区间重叠/交叉重复问题
for(int j = i; j<=n; j++) ms.erase(ms.find(s[j] - s[i-1]));
}
cout << ans << endl;
return 0;
}
/* test samples
5
10 9 8 12 14
*/

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#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
n++;
cout << n << endl;
return 0;
}

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#include<bits/stdc++.h>
using namespace std;
typedef struct LNode{
int data;
struct LNode *next;
}LNode, *LinkList;
bool InitList(LinkList &L){
L = (LNode *)malloc(sizeof(LNode)); // 带头节点
if(L == NULL) return false;
L->next = NULL;
return true;
}
void InputList(LinkList &L){
int x; cin>>x;
LinkList tail = L;
while(x != -99999){ // 输入 -99999 结束输入
LNode *newNode = (LNode*)malloc(sizeof(LNode));
newNode->data = x; newNode->next = NULL;
tail->next = newNode;
tail = newNode;
cin >> x;
}
}
void PrintList(LinkList &L){
LinkList p = L->next;
while(p != NULL){
cout << p->data << ",\n"[p->next == NULL];
p = p->next;
}
}
LinkList p1, p2, p3, tmp;
void MergeList1(LinkList &L1, LinkList &L2, LinkList &L3){
// 升序合并并去重
p1 = L1->next; p2 = L2->next;
L3 = p3 = L1;
while(p1 && p2){
if(p1->data < p2->data){
p3->next = p1; p3 = p1; p1 = p1->next;
}else if(p2->data < p1->data){
p3->next = p2; p3 = p2; p2 = p2->next;
}else{
p3->next = p1; p3 = p1; p1 = p1->next;
tmp = p2->next; delete p2; p2 = tmp;
}
}
p3->next = p1?p1:p2;
delete L2;
}
void MergeList2(LinkList &L1, LinkList &L2, LinkList &L3){
// 降序合并
p1 = L1->next; p2 = L2->next;
L3 = L1;
L3->next = NULL;
while(p1 || p2){
if(!p1) { tmp = p2; p2 = p2->next; }
else if(!p2) { tmp = p1; p1 = p1->next; }
else if(p1->data <= p2->data) { tmp = p1; p1 = p1->next; }
else { tmp = p2; p2 = p2->next; }
tmp->next = L3->next; L3->next = tmp;
}
delete L2;
}
void MixList(LinkList &L1, LinkList &L2, LinkList &L3){
// 两链表相交
p1 = L1->next; p2 = L2->next;
L3 = p3 = L1;
while(p1 && p2){
if(p1->data == p2->data){
p3->next = p1; p3 = p1; p1 = p1->next;
tmp = p2; p2 = p2->next; delete tmp;
}else if(p1->data < p2->data){
tmp = p1; p1 = p1->next; delete tmp;
}else{
tmp = p2; p2 = p2->next; delete tmp;
}
}
while(p1) { tmp = p1; p1 = p1->next; delete tmp; }
while(p2) { tmp = p2; p2 = p2->next; delete tmp; }
p3->next = NULL;
delete L2;
}
void DiffList(LinkList &L1, LinkList &L2, int &n){
// 两链表相差
p1 = L1->next; p2 = L2->next;
p3 = L1;
while(p1 && p2){
if(p1->data < p2->data){
p3 = p1; p1 = p1->next; n++;
}else if(p1->data > p2->data){
p2 = p2->next;
}else{
p3->next = p1->next;
tmp = p1; p1 = p1->next; delete tmp;
}
}
}
int main(){
LinkList L1, L2, L3; InitList(L1); InitList(L2); InitList(L3);
InputList(L1); InputList(L2); // PrintList(L1); PrintList(L2); // 初始化并输入输出两个链表
MergeList1(L1, L2, L3);
cout << "两链表升序合并并去重: "; PrintList(L3);
LinkList L4, L5, L6; InitList(L4); InitList(L5); InitList(L6);
InputList(L4); InputList(L5); // PrintList(L4); PrintList(L5); // 初始化并输入输出两个链表
MergeList2(L4, L5, L6);
cout << "两链表降序合并: "; PrintList(L6);
LinkList L7, L8, L9; InitList(L7); InitList(L8); InitList(L9);
InputList(L7); InputList(L8);
MixList(L7, L8, L9);
cout << "两链表相交: "; PrintList(L9);
LinkList L10, L11; InitList(L10); InitList(L11);
InputList(L10); InputList(L11); int n = 0;
DiffList(L10, L11, n);
cout << "两链表相差: "; PrintList(L10);
cout << "元素个数位: " << n << endl;
return 0;
}
/* test samples
1 3 5 7 9 13 15 -99999
2 4 6 8 10 11 13 14 15 -99999
*/

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// 主元素寻找
#include <iostream>
#include <vector>
using namespace std;
int findMajorityElement(const vector<int>& nums) {
int candidate = -1;
int count = 0;
// 第一次遍历：找出候选主元素
for (int num : nums) {
if (count == 0) {
candidate = num;
}
count += (num == candidate) ? 1 : -1;
}
// 第二次遍历：验证候选主元素
count = 0;
for (int num : nums) {
if (num == candidate) {
count++;
}
}
// 如果候选元素出现次数超过数组长度的一半，则为所求主元素
if (count > nums.size() / 2) {
return candidate;
} else {
return -1;
}
}
int main() {
vector<int> nums = {0, 5, 3, 5, 5, 7, 5, 5};
int result = findMajorityElement(nums);
if (result != -1) {
cout << "主元素是: " << result << endl;
} else {
cout << "没有主元素" << endl;
}
return 0;
}

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#include<bits/stdc++.h>
using namespace std;
vector<vector<int>> results;
void backtrack(vector<int>& current, vector<bool>& used, int n){
if(current.size() == n){
results.push_back(current);
return;
}
for(int i = 1; i <= n; i++){
if(!used[i]){
current.push_back(i);
used[i] = true;
backtrack(current, used, n);
current.pop_back();
used[i] = false;
}
}
}
int main(){
int n; cin >> n;
vector<int> current;
vector<bool> used(n+1, false);
backtrack(current, used, n);
for(const auto& perm:results){
for(int num:perm){
cout << num << " ";
}
cout << endl;
}
return 0;
}

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#include<bits/stdc++.h>
using namespace std;
const int N = 1e2 + 10; // 定义最大数组大小
int a[N]; // 存储排列结果
bool vis[N]; // 标记数字是否被访问过
int n; // 排列的大小
// 回溯函数，dep表示当前处理到第几个数字
void dfs(int dep) {
if (dep == n + 1) { // 如果已经处理完所有数字
for (int i = 1; i <= n; i++) { // 输出排列
cout << a[i] << " \n"[i==n];
}
return;
}
for (int i = 1; i <= n; i++) { // 尝试将1到n的每个数字放入当前位置
if (vis[i]) continue; // 如果该数字已经被使用，跳过
vis[i] = true; // 标记该数字为已使用
a[dep] = i; // 将数字放入当前位置
dfs(dep + 1); // 递归处理下一个位置
vis[i] = false; // 回溯，恢复状态
}
}
int main() {
cin >> n; // 输入排列的大小
dfs(1); // 从第一个位置开始递归
return 0;
}

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// lanqiao1508 N皇后问题
#include<bits/stdc++.h>
using namespace std;
const int N = 15;
int n, ans;
int vis[N][N];
void dfs(int dep){
if(dep == n + 1){
ans++;
return;
}
for(int i = 1; i <= n; i++){ // 为什么下面只标记同一行，因为每次搜寻时只会在同一行(深度)寻找一个皇后
if(vis[dep][i]) continue; // 所以只需要排除某行某个元素的同一列即可
// 修改状态
for(int _i = 1; _i <= n; _i++) vis[_i][i]++; // 标记同一列
// 表壳 X 形模版
for(int _i = dep, _j = i; _i >= 1 && _j >= 1; _i--, _j--) vis[_i][_j]++; // 标记左上对角线
for(int _i = dep, _j = i; _i <= n && _j >= 1; _i++, _j--) vis[_i][_j]++; // 标记右上对角线
for(int _i = dep, _j = i; _i >= 1 && _j <= n; _i--, _j++) vis[_i][_j]++; // 标记左下对角线
for(int _i = dep, _j = i; _i <= n && _j <= n; _i++, _j++) vis[_i][_j]++; // 标记右下对角线
// 搜查下一深度
dfs(dep + 1);
// 恢复现场
for(int _i = 1; _i <= n; _i++) vis[_i][i]--;
for(int _i = dep, _j = i; _i >= 1 && _j >= 1; _i--, _j--) vis[_i][_j]--;
for(int _i = dep, _j = i; _i <= n && _j >= 1; _i++, _j--) vis[_i][_j]--;
for(int _i = dep, _j = i; _i >= 1 && _j <= n; _i--, _j++) vis[_i][_j]--;
for(int _i = dep, _j = i; _i <= n && _j <= n; _i++, _j++) vis[_i][_j]--;
}
}
int main(){
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> n;
dfs(1);
cout << ans << '\n';
return 0;
}
/* test samples
5
*/

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// lanqiao182 小朋友崇拜圈
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, a[N], dfn[N], idx, mindfn;
int dfs(int x){
dfn[x] = ++idx;
if(dfn[a[x]]){
if(dfn[a[x]] >= mindfn) return dfn[x] - dfn[a[x]] + 1;
return 0;
}
return dfs(a[x]);
}
int main(){
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> n;
for(int i = 1; i <= n; i++) cin >> a[i];
int ans = 0;
for(int i = 1; i <= n; i++){
if(!dfn[i]){
mindfn = idx + 1;
ans = max(ans, dfs(i));
}
}
cout << ans << endl;
return 0;
}
/* test samples
9
3 4 2 5 3 8 4 6 9
*/

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// lanqiao178 全球变暖
#include<bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
char mp[N][N];
int n, scc, col[N][N];
bool vis[N*N];
// 移动向量
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
void dfs(int x, int y){
col[x][y] = scc;
for(int i = 0; i < 4; i++){
int nx = x + dx[i], ny = y + dy[i];
if( col[nx][ny] || mp[nx][ny] == '.') continue; // 不需要判断nx, ny是否在地图里面，因为地图边缘都是海洋
dfs(nx ,ny);
}
}
int main(){
ios::sync_with_stdio, cin.tie(0), cout.tie(0);
cin >> n;
for(int i = 1; i <= n; i++) cin >> mp[i] + 1;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
if(col[i][j] || mp[i][j] == '.') continue;
scc++; // scc表示当前颜色编号
dfs(i, j);
}
}
int ans = 0;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
if(mp[i][j] == '.') continue;
bool tag = true;
for(int k = 0; k < 4; k++){
int x = i + dx[k], y = j + dy[k];
if(mp[x][y] == '.') tag = false;
}
if(tag){
if(!vis[col[i][j]]) ans++;
vis[col[i][j]] = true;
}
}
}
cout << scc - ans << '\n';
return 0;
}
/* test samples
7
.......
.##....
.##....
....##.
..####.
...###.
.......
*/

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// lanqiao2942 数字王国之军训排队
#include<bits/stdc++.h>
using namespace std;
const int N = 15;
int a[N], n;
vector<int> v[N];
bool dfs(int cnt, int dep){
if(dep == n+1){
// 检查当前方案的合法性
for(int i = 1; i <= cnt; i++){
for(int j = 0; j < v[i].size(); j++){
for(int k = j+1; k < v[i].size(); k++){
if(v[i][k] % v[i][j] == 0) return false;
}
}
}
return true;
}
// 枚举每个人所属的队伍
for(int i = 1; i <= cnt; i++){
v[i].push_back(a[dep]);
if(dfs(cnt, dep + 1)) return true;
// 恢复现场
v[i].pop_back();
}
return false;
}
int main(){
cin >> n;
for(int i = 1; i <= n; i++) cin >> a[i];
sort(a+1, a+1+n);
for(int i = 1; i <= n; i++){
if(dfs(i, 1)){
cout << i << '\n';
break;
}
}
return 0;
}
/* test samples
4
2 3 4 4
*/

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// lanqiao2942 数字王国之军训排队
#include<bits/stdc++.h>
using namespace std;
const int N = 15;
int a[N], n;
vector<int> v[N];
// 剪枝
bool dfs(int cnt, int dep){
if(dep == n+1){
return true;
}
// 枚举每个人所属的队伍
for(int i = 1; i <= cnt; i++){
bool tag = true;
for(const auto &j:v[i]){
if(a[dep] % j == 0){
tag = false;
break;
}
}
if(!tag) continue;
v[i].push_back(a[dep]);
if(dfs(cnt, dep + 1)) return true;
// 恢复现场
v[i].pop_back();
}
return false;
}
int main(){
cin >> n;
for(int i = 1; i <= n; i++) cin >> a[i];
sort(a+1, a+1+n);
for(int i = 1; i <= n; i++){
if(dfs(i, 1)){
cout << i << '\n';
break;
}
}
return 0;
}
/* test samples
4
2 3 4 4
*/

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// 蓝桥 3008 特殊的三角形
#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+9;
int cnt[N], prefix[N];
void dfs(int dep, int st, int mul, int sum){
if(mul > 1e6) return;
if(dep == 4){
cnt[mul]++;
return;
}
int up = pow(1e6 / mul, 1.0/(4-dep)) + 3;
for(int i = st + 1; i < (dep == 3?sum:up); i++){
dfs(dep+1, i, mul * i, sum + i);
}
}
int main(){
dfs(1, 0, 1, 0);
for(int i = 1; i <= 1e6; i++){
prefix[i] = prefix[i-1] + cnt[i];
}
int q; cin >> q;
while(q--){
int l, r; cin >> l >> r;
cout << prefix[r] - prefix[l-1] << '\n';
}
return 0;
}
/* test samples
4
1 10
30 50
60 200
200 400
*/

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// 蓝桥3075 特殊的多边形
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+9;
int n, cnt[N], prefix[N];
void dfs(int dep, int st, int mul, int sum){
// 剪枝1
if(mul > 1e5) return;
if(dep == n + 1){
cnt[mul]++;
return;
}
// 剪枝2
int up = pow(1e5 / mul, 1.0/(n - dep + 1)) + 3;
for(int i = st + 1; i < (dep == n?min(sum, up):up); i++){
dfs(dep+1, i, mul * i, sum + i);
}
}
int main(){
int q; cin >> q >> n;
dfs(1, 0, 1, 0);
for(int i = 1; i <= 1e5; i++){
prefix[i] = prefix[i-1] + cnt[i];
}
while(q--){
int l, r; cin >> l >> r;
cout << prefix[r] - prefix[l-1] << '\n';
}
return 0;
}
/* test samples
4 3
1 10
30 50
60 200
200 400
*/

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// 斐波那契数递归优化
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const ll p = 1e9+7;
const int inf = 1e9, N = 1e5+3;
ll dp[N];
ll f(int n){
if( n <= 2) return 1;
if(dp[n] != -1) return dp[n];
return dp[n] = (f(n -1 ) + f(n - 2)) % p;
}
int main(){
memset(dp, -1, sizeof dp);
int n; cin >> n;
cout << f(n) << endl;
return 0;
}
/* test samples 976496506
5000
*/

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#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const ll p = 1e9 + 7;
const int inf = 1e9, N = 1e3 + 3;
int n, m, k, sx, sy, fx, fy, h[N][N];
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
int dp[N][N][2];
bool inmp(int x, int y){
return x >= 1 && x <= n && y >= 1 && y <= m;
}
// 返回值表示能否到达终点(fx, fy), t表示当前使用的喷气背包的次数
bool dfs(int x, int y, int t){
if(x == fx && y == fy) return true;
if(dp[x][y][t] != -1) return dp[x][y][t];
for(int i = 0; i < 4; i++){
int nx = x + dx[i], ny = y + dy[i];
if(!inmp(nx, ny)) continue;
if(!t){ // 没用过背包
// 不用
if(h[x][y] > h[nx][ny] && dfs(nx, ny, 0)) return dp[x][y][t] = true;
// 用
if(h[x][y] + k > h[nx][ny] && dfs(nx, ny, 1)) return dp[x][y][t] = true;
}else{ // 已经用过一次背包了
// 不用
if(h[x][y] > h[nx][ny] && dfs(nx, ny, 1)) return dp[x][y][t] = true;
}
}
return dp[x][y][t] = false;
}
int main(){
memset(dp, -1, sizeof dp);
cin >> n >> m >> k;
cin >> sx >> sy >> fx >> fy;
for(int i = 1; i <= n; i++){
for(int j = 1; j <= m; j++){
cin >> h[i][j];
}
}
cout << (dfs(sx, sy, 0)?"Yes":"No") << '\n';
return 0;
}
/* test samples 1 -> YES
5 5 30
1 1 5 5
3 20 13 12 11
19 17 33 72 10
12 23 12 23 9
21 43 23 12 2
21 34 23 12 1
*/
/* test samples 1 -> NO
5 5 10
1 1 5 5
3 2 13 12 11
1 17 33 72 10
12 23 12 23 9
21 43 23 12 2
21 34 23 12 1
*/