'commit'

TangDou/LanQiaoBei/LanQiao14STEMA202208/2.cpp

@@ -0,0 +1,17 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+int cnt;
+/*
+110
+10
+*/
+int main() {
+    int n;
+    cin >> n;
+    for (int i = 100; i <= n; i++)
+        if (i % 10 != 3 && i % 100 / 10 != 5 && i % 1000 / 100 != 7)
+            cnt++;
+    printf("%d\n", cnt);
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/24.cpp

@@ -0,0 +1,70 @@
+/* HDU1427 速算24点 */
+
+#include <iostream>
+#include <algorithm>
+#include <stdio.h>
+#include <string.h>
+
+using namespace std;
+
+const int N = 4;
+int a[N];
+bool flag;
+
+bool dfs(int cnt, int result1, int result2) {
+    if (cnt == N - 1) {
+        if (result1 + result2 == 24 || result1 - result2 == 24 || result1 * result2 == 24)
+            flag = true;
+        if (result2 != 0 && result1 % result2 == 0 && result1 / result2 == 24)
+            flag = true;
+    } else if (!flag) {
+        // 先算前2项
+        dfs(cnt + 1, result1 + result2, a[cnt + 1]);
+        dfs(cnt + 1, result1 - result2, a[cnt + 1]);
+        dfs(cnt + 1, result1 * result2, a[cnt + 1]);
+        if (result2 != 0 && result1 % result2 == 0)
+            dfs(cnt + 1, result1 / result2, a[cnt + 1]);
+
+        // 先算后2项，相当与后2项计算是加括号的
+        dfs(cnt + 1, result1, result2 + a[cnt + 1]);
+        dfs(cnt + 1, result1, result2 - a[cnt + 1]);
+        dfs(cnt + 1, result1, result2 * a[cnt + 1]);
+        if (a[cnt + 1] != 0 && result2 % a[cnt + 1] == 0)
+            dfs(cnt + 1, result1, result2 / a[cnt + 1]);
+    }
+    return false;
+}
+
+int main() {
+    for (;;) {
+        char s[4];
+        for (int i = 0; i < N; i++) {
+            if (scanf("%s", s) == EOF)
+                return 0;
+            if (strlen(s) == 2)
+                a[i] = 10;
+            else if (s[0] == 'A')
+                a[i] = 1;
+            else if (s[0] == 'J')
+                a[i] = 11;
+            else if (s[0] == 'Q')
+                a[i] = 12;
+            else if (s[0] == 'K')
+                a[i] = 13;
+            else
+                a[i] = s[0] - '0';
+        }
+
+        sort(a, a + N);
+        flag = false;
+        do {
+            dfs(1, a[0], a[1]);
+            if (flag)
+                break;
+        } while (next_permutation(a, a + N));
+
+        printf("%s\n", flag ? "Yes" : "No");
+    }
+
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/3_0.cpp

@@ -0,0 +1,18 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+string str = "cacd";
+// 3
+int bucket[26];
+int cnt;
+int main() {
+    //桶计数
+    for (int i = 0; i < str.size(); i++) bucket[str[i] - 'a']++;
+    for (int i = 0; i < 26; i++)
+        if (bucket[i]) cnt++;
+
+    cout << cnt << endl;
+
+    // t.size()!=cnt
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/3_1.cpp

@@ -0,0 +1,32 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 26;
+unordered_set<char> S;
+bool st[N];
+string res;
+string s;
+void dfs(int u, string t) {
+    if (u == s.size()) {               //走完全程
+        if (t.size() == S.size()) {    //是不是本路径取得的是最小字串的字符串长度，比如是不是4
+            if (res == "")             //如果是第一次
+                res = t;               // t就成为大王
+            else if (t < res)          //否则t需要击败现在的大王res
+                res = t;
+        }
+        return;
+    }
+    if (!st[s[u] - 'a']) {      //如果当前字符没有在本路径中出现过，则可以尝试用上当前的字符
+        st[s[u] - 'a'] = true;  //标识已用过
+        dfs(u + 1, t + s[u]);   //下一个位置
+        st[s[u] - 'a'] = false; //回溯
+    }
+    dfs(u + 1, t); //放弃当前位置
+}
+int main() {
+    cin >> s;
+    for (int i = 0; i < s.size(); i++) S.insert(s[i]); //去重后的最小子串长度为S.size()
+    dfs(0, "");
+    cout << res << endl;
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/3_2.cpp

@@ -0,0 +1,45 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int N = 26;
+// acdaccdd
+unordered_set<char> S; //有哪些字符
+bool st[N];            //是不是使过了
+string res;            //结果，字典序最小
+string s;              //原始串
+string t;
+void dfs(int u) {
+    //出口
+    if (u == s.size()) {
+        if (t.size() == S.size())
+            if (t < res) res = t; // t<res
+        return;
+    }
+    //没有使过
+    if (!st[s[u] - 'a']) {
+        //使上
+        st[s[u] - 'a'] = true;
+        t.push_back(s[u]);
+        //下一个
+        dfs(u + 1);
+        //回溯
+        t.pop_back();
+        st[s[u] - 'a'] = false;
+    }
+    //使过或者没使过但不想使，都走这条路径
+    dfs(u + 1);
+}
+
+int main() {
+    //初始化极大值
+    for (int i = 0; i < 200; i++) res = res + 'z';
+    //读入字符串
+    cin >> s;
+    //去重形成一个集合
+    for (int i = 0; i < s.size(); i++) S.insert(s[i]);
+    //深搜
+    dfs(0);
+    //输出
+    cout << res << endl;
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/4_dfs_int.cpp

@@ -0,0 +1,35 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int INF = 0x3f3f3f3f;
+
+const int N = 110;
+int a[N][N];
+int n, m;
+
+// dfs解法
+/*
+2 3
+1 3 5
+2 4 6
+
+答案：
+13
+*/
+
+int dfs(int x, int y) {
+    if (x == n && y == m) return a[n][m];
+    if (x + 1 <= n && y + 1 <= m) return a[x][y] + min(dfs(x + 1, y), dfs(x, y + 1));
+    if (x + 1 <= n && y + 1 > m) return a[x][y] + dfs(x + 1, y);
+    if (x + 1 > n && y + 1 <= m) return a[x][y] + dfs(x, y + 1);
+}
+
+int main() {
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++)
+            cin >> a[i][j];
+
+    printf("%d\n", dfs(1, 1));
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/4_dfs_int_jiyihua.cpp

@@ -0,0 +1,36 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int INF = 0x3f3f3f3f;
+
+const int N = 110;
+int a[N][N];
+int n, m;
+int f[N][N];
+
+// dfs解法
+/*
+2 3
+1 3 5
+2 4 6
+
+答案：
+13
+*/
+
+int dfs(int x, int y) {
+    if (f[x][y]) return f[x][y];
+    if (x > n || y > m) return INF;
+    if (x == n && y == m) return a[n][m];
+    return f[x][y] = a[x][y] + min(dfs(x + 1, y), dfs(x, y + 1));
+}
+
+int main() {
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++)
+            cin >> a[i][j];
+
+    printf("%d\n", dfs(1, 1));
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/4_dfs_void_1.cpp

@@ -0,0 +1,46 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int INF = 0x3f3f3f3f;
+
+const int N = 110;
+int a[N][N];
+int n, m;
+int res = INF;
+// dfs解法
+/*
+2 3
+1 3 5
+2 4 6
+
+
+答案：
+13
+
+2 3
+1 2 3
+4 5 6
+*/
+int dx[] = {1, 0};
+int dy[] = {0, 1};
+void dfs(int x, int y, int sum) {
+    if (x == n && y == m) {
+        res = min(res, sum);
+        return;
+    }
+    for (int i = 0; i < 2; i++) {
+        int tx = x + dx[i], ty = y + dy[i];
+        if (tx > n || ty > m) continue;
+        dfs(tx, ty, sum + a[tx][ty]);
+    }
+}
+int main() {
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++)
+            cin >> a[i][j];
+
+    dfs(1, 1, a[1][1]);
+    printf("%d\n", res);
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/4_dfs_void_2.cpp

@@ -0,0 +1,39 @@
+#include <cstdio>
+#include <cstring>
+#include <algorithm>
+#include <iostream>
+#include <cmath>
+
+using namespace std;
+const int INF = 0x3f3f3f3f;
+const int N = 110;
+int a[N][N];
+int n, m;
+int res = INF;
+
+void dfs(int x, int y, int sum) {
+    if (x == n && y == m) {
+        res = min(res, sum);
+        return;
+    }
+    if (x + 1 <= n)
+        dfs(x + 1, y, sum + a[x + 1][y]);
+    if (y + 1 <= m)
+        dfs(x, y + 1, sum + a[x][y + 1]);
+}
+/*
+2 3
+1 3 5
+2 4 6
+
+13
+*/
+int main() {
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++)
+            cin >> a[i][j];
+    dfs(1, 1, a[1][1]);
+    printf("%d\n", res);
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/4_dp.cpp

@@ -0,0 +1,39 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+const int INF = 0x3f3f3f3f;
+
+const int N = 110;
+int n, m;
+int a[N][N];
+int res = INF;
+int f[N][N];
+/*
+动态规划
+2 3
+1 3 5
+2 4 6
+
+答案：
+13
+*/
+
+int main() {
+    memset(f, 0x3f, sizeof f);
+
+    cin >> n >> m;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++)
+            cin >> a[i][j];
+
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            if (i == 1 && j == 1)
+                f[i][j] = a[i][j];
+            else
+                f[i][j] = min(f[i - 1][j], f[i][j - 1]) + a[i][j];
+        }
+
+    printf("%d\n", f[n][m]);
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/5_dfs.cpp

@@ -0,0 +1,39 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int N = 2010;
+int f[N][N];
+
+char a[N], b[N];
+int al, bl;
+/*
+测试用例：
+abcd
+ebde
+
+答案：3
+*/
+int dfs(int i, int j) {
+    if (i == 0) return j;
+    if (j == 0) return i;
+
+    if (~f[i][j]) return f[i][j];
+    
+    //增加、删除
+    f[i][j] = min(dfs(i - 1, j) + 1, dfs(i, j - 1) + 1);
+
+    //修改
+    if (a[i] == b[j])
+        return f[i][j] = min(f[i][j], dfs(i - 1, j - 1));
+    else
+        return f[i][j] = min(f[i][j], dfs(i - 1, j - 1) + 1);
+}
+
+int main() {
+    memset(f, -1, sizeof f);
+    cin >> a + 1 >> b + 1;
+    al = strlen(a + 1), bl = strlen(b + 1);
+    printf("%d\n", dfs(al, bl));
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/5_dp.cpp

@@ -0,0 +1,32 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int N = 2010;
+char a[N], b[N];
+int al, bl;
+int f[N][N];
+
+int main() {
+    cin >> a + 1 >> b + 1;
+    al = strlen(a + 1), bl = strlen(b + 1);
+
+    //初始化
+    for (int i = 1; i <= bl; i++) f[0][i] = i;
+    for (int i = 1; i <= al; i++) f[i][0] = i;
+
+    for (int i = 1; i <= al; i++)
+        for (int j = 1; j <= bl; j++) {
+            //增加删除
+            f[i][j] = min(f[i - 1][j] + 1, f[i][j - 1] + 1);
+            //修改
+            if (a[i] == b[j])
+                f[i][j] = min(f[i][j], f[i - 1][j - 1]);
+            else
+                f[i][j] = min(f[i][j], f[i - 1][j - 1] + 1);
+        }
+
+    printf("%d\n", f[al][bl]);
+
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/TD.cpp

@@ -0,0 +1,29 @@
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int N = 2010;
+
+int f[N][N], al, bl;
+char a[N], b[N];
+
+int main() {
+    cin >> a + 1 >> b + 1;
+    al = strlen(a + 1), bl = strlen(b + 1);
+
+    for (int i = 1; i <= bl; i++) f[0][i] = i;
+    for (int i = 1; i <= al; i++) f[i][0] = i;
+
+    for (int i = 1; i <= al; i++)
+        for (int j = 1; j <= bl; j++) {
+            f[i][j] = min(f[i - 1][j] + 1, f[i][j - 1] + 1);
+
+            if (a[i] == b[j])
+                f[i][j] = min(f[i][j], f[i - 1][j - 1]);
+            else
+                f[i][j] = min(f[i][j], f[i - 1][j - 1] + 1);
+        }
+
+    printf("%d\n", f[al][bl]);
+    return 0;
+}

TangDou/LanQiaoBei/LanQiao14STEMA202208/编程解决小学奥数问题(竖式).md

@@ -0,0 +1,145 @@
+### 一、题目大意
+昨日，糖豆做了一道小学四年级数学举一反三试题：
+
+$\large ~~~~abcdef$
+$\large +  abcdef$
+$\large ----$
+$\large ~~~jeghba$
+
+已知题目中所有字符是非负整数,求$abcdef$和$jeghba$的值是多少？
+
+此题我给出的数学方法是画树形图，标注分支，然后尝试每个分支，判断是否符合题目，是否可以剪枝，但计算量很大，糖豆历时$20$分钟，完成此题。
+
+回头思考，既然都画了树形图，不如使用编程的办法尝试一下，随即做了两套办法，现整理给同学们参考，望大家深入理解深搜的优势，以及它的实现细节。
+
+### 二、循环暴力法
+
+$Running~time:1000ms$
+$1$秒内完成计算，无法优化了，没法剪枝。贵在思路简单，好上手。
+```c++
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int N = 10;
+int a[N], al; //用于将最终的两个原数字相加的结果 拆分到数组中
+int b[N], bl; //用于拆分原数字到数组中
+
+int main() {
+    //记录开始时间
+    clock_t begin = clock();
+
+    // 1234
+    //排重后数字个数
+    unordered_set<int> _set1, _set2;
+    
+    for (int i = 123456; i < 1000000 / 2; i++) {
+        int x = i;
+        int y = i + i;
+
+        al = 0;
+        while (y) a[++al] = y % 10, y /= 10;
+
+        bl = 0;
+        while (x) b[++bl] = x % 10, x /= 10;
+        
+        _set1.clear();
+        for (int j = 1; j <= al; j++) _set1.insert(a[j]);
+        if (_set1.size() != 6) continue;
+
+        _set2.clear();
+        for (int j = 1; j <= bl; j++) _set2.insert(b[j]);
+        if (_set2.size() != 6) continue;
+
+        if (a[1] == b[6] && a[2] == b[5] && a[5] == b[2]
+            && !_set2.count(a[3]) && !_set2.count(a[4]) && !_set2.count(a[6]))
+            printf("%d\n", i);
+    }
+
+    //记录结束时间
+    clock_t end = clock();
+    cout << "Running time:" << (double)(end - begin) / CLOCKS_PER_SEC * 1000 << "ms" << endl;
+
+    return 0;
+}
+```
+
+### 三、深搜+剪枝
+$Running~~ time:3ms$
+
+逆天的$3ms$！整整快了$300$倍~
+
+<font color='red' size=4><b>分析：
+因深搜可以在行进过程中随时发现问题，随时终止此分支前进，防止了很多无用的枚举。
+</b></font>
+
+```c++
+#include <bits/stdc++.h>
+
+using namespace std;
+/*
+
+输出：
+468532
+*/
+const int N = 10;
+int num;      //不断增加的前缀，最终是选择的数字
+bool st[N];   // 0~9哪个数字使用过
+int a[N], al; //用于将最终的两个原数字相加的结果 拆分到数组中
+int b[N], bl; //用于拆分原数字到数组中
+
+void dfs(int u) {
+    if (u == 7) { //完成所有数位的枚举
+        int sum = num << 1, tmp = num;
+
+        //拆分到数组中
+        al = 0;
+        while (sum) a[++al] = sum % 10, sum /= 10;
+
+        bl = 0;
+        while (tmp) b[++bl] = tmp % 10, tmp /= 10;
+
+        //判断条件,能剪枝的代码，尽量提前，减少后面无用的计算
+        // 放在unordered_set前面，可以做到: Running time:3ms
+        if (a[1] != b[6] || a[2] != b[5] || a[5] != b[2]) return;
+        if (st[a[6]] || st[a[4]] || st[a[3]]) return;
+
+        // Running time:526ms
+        //判断是不是结果存在数位值重复
+        unordered_set<int> _set;
+        for (int i = 1; i <= al; i++) _set.insert(a[i]);
+        if (_set.size() < 6) return;
+
+        //全部符合要求，输出结果
+        printf("%d\n", num);
+        return;
+    }
+    for (int i = 0; i <= 9; i++) {
+        //首位不能达到5，否则就是7位了,剪枝
+        //如果是首位，a!=1,因为加法的结果个数是1，加法结果最小应该是2开始
+        // a只能是2,4
+        if (u == 1 && (i != 2 && i != 4)) continue;
+        if (!st[i]) {
+            st[i] = true;
+            num = num * 10 + i; //增加上此位
+            dfs(u + 1);         //下一个位置
+            // num = (num - i) / 10; //减回去此位
+            //简写
+            num = num / 10; //减回去此位
+            st[i] = false;
+        }
+    }
+}
+int main() {
+    //记录开始时间
+    clock_t begin = clock();
+
+    dfs(1);
+
+    //记录结束时间
+    clock_t end = clock();
+    cout << "Running time:" << (double)(end - begin) / CLOCKS_PER_SEC * 1000 << "ms" << endl;
+
+    return 0;
+}
+```