Initial commit

2025-03-14 22:39:32 +08:00
commit 7b1e0f329e
38 changed files with 1153 additions and 0 deletions
--- a/.DS_Store
+++ b/.DS_Store
--- a/14lanqiao/test1.cpp
+++ b/14lanqiao/test1.cpp
@@ -0,0 +1,28 @@
+#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;
+}
--- a/14lanqiao/test2.cpp
+++ b/14lanqiao/test2.cpp
@@ -0,0 +1,13 @@
+#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;
+}
--- a/14lanqiao/test3-1.cpp
+++ b/14lanqiao/test3-1.cpp
@@ -0,0 +1,49 @@
+#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
+*/
--- a/14lanqiao/test3-2.cpp
+++ b/14lanqiao/test3-2.cpp
@@ -0,0 +1,26 @@
+#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
+*/
--- a/14lanqiao/test4.cpp
+++ b/14lanqiao/test4.cpp
@@ -0,0 +1,50 @@
+#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
+*/
--- a/14lanqiao/test5-1.cpp
+++ b/14lanqiao/test5-1.cpp
@@ -0,0 +1,33 @@
+#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
+*/
--- a/14lanqiao/test5-2.cpp
+++ b/14lanqiao/test5-2.cpp
@@ -0,0 +1,25 @@
+#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
+*/
--- a/15lanqiao/test1-1.cpp
+++ b/15lanqiao/test1-1.cpp
@@ -0,0 +1,15 @@
+#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)
+*/
--- a/15lanqiao/test1-2.cpp
+++ b/15lanqiao/test1-2.cpp
@@ -0,0 +1,17 @@
+#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
+
+*/
--- a/15lanqiao/test2.cpp
+++ b/15lanqiao/test2.cpp
@@ -0,0 +1,19 @@
+#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)
+*/
--- a/15lanqiao/test3.cpp
+++ b/15lanqiao/test3.cpp
@@ -0,0 +1,30 @@
+#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
+*/
--- a/15lanqiao/test4.cpp
+++ b/15lanqiao/test4.cpp
@@ -0,0 +1,34 @@
+#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
+*/
--- a/15lanqiao/test5.cpp
+++ b/15lanqiao/test5.cpp
@@ -0,0 +1,45 @@
+#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。
+*/
--- a/15lanqiao/test6.cpp
+++ b/15lanqiao/test6.cpp
@@ -0,0 +1,57 @@
+#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
+*/
--- a/15lanqiao/test7.cpp
+++ b/15lanqiao/test7.cpp
@@ -0,0 +1,26 @@
+#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
+*/
--- a/15lanqiao/test8.cpp
+++ b/15lanqiao/test8.cpp
@@ -0,0 +1,44 @@
+#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
+*/
--- a/test100.cpp
+++ b/test100.cpp
@@ -0,0 +1,9 @@
+#include<bits/stdc++.h>
+using namespace std;
+
+int main(){
+    int n;
+    n++;
+    cout << n << endl;
+    return 0;
+}
--- a/test99-1.cpp
+++ b/test99-1.cpp
@@ -0,0 +1,134 @@
+#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
+*/
--- a/test99-2.cpp
+++ b/test99-2.cpp
@@ -0,0 +1,44 @@
+// 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;
+}
--- a/test99-3-1.cpp
+++ b/test99-3-1.cpp
@@ -0,0 +1,34 @@
+#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;
+}
--- a/test99-3-2.cpp
+++ b/test99-3-2.cpp
@@ -0,0 +1,31 @@
+#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;
+}
--- a/test99-3-3.cpp
+++ b/test99-3-3.cpp
@@ -0,0 +1,44 @@
+// 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
+*/
--- a/BIN
+++ b/BIN
--- a/test99-3-4.cpp
+++ b/test99-3-4.cpp
@@ -0,0 +1,33 @@
+// 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
+*/
--- a/BIN
+++ b/BIN
--- a/test99-3-5.cpp
+++ b/test99-3-5.cpp
@@ -0,0 +1,59 @@
+// 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
+.......
+.##....
+.##....
+....##.
+..####.
+...###.
+.......
+*/
--- a/BIN
+++ b/BIN
--- a/test99-4-1-1.cpp
+++ b/test99-4-1-1.cpp
@@ -0,0 +1,44 @@
+// 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
+*/
--- a/test99-4-1-2.cpp
+++ b/test99-4-1-2.cpp
@@ -0,0 +1,45 @@
+// 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
+*/
--- a/BIN
+++ b/BIN
--- a/test99-4-2.cpp
+++ b/test99-4-2.cpp
@@ -0,0 +1,37 @@
+// 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
+*/
--- a/BIN
+++ b/BIN
--- a/test99-4-3.cpp
+++ b/test99-4-3.cpp
@@ -0,0 +1,39 @@
+// 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
+*/
--- a/BIN
+++ b/BIN
--- a/test99-4-4.cpp
+++ b/test99-4-4.cpp
@@ -0,0 +1,23 @@
+// 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 
+*/ 
--- a/BIN
+++ b/BIN
--- a/test99-4-5.cpp
+++ b/test99-4-5.cpp
@@ -0,0 +1,66 @@
+#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
+*/