Initial commit

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;
+}

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;
+}

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
+*/

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
+*/

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
+*/

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
+*/

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
+*/