feat: 新增统计特殊字母数量的算法实现及相关文档

Author: ttkqwe123

CCF-CSP/ [CSPro 31] 坐标变换（其二）.cpp

@@ -0,0 +1,51 @@
+/*
+ * @Author: tkzzzzzz6
+ * @Date: 2026-05-26 22:11:02
+ * @LastEditors: tkzzzzzz6
+ * @LastEditTime: 2026-05-26 22:52:41
+ */
+#include <cmath>
+#include <cstdio>
+#include <iomanip>
+#include <iostream>
+
+using namespace std;
+const int N = 1e5 + 10;
+double op1[N], op2[N];
+
+int main() {
+    int n, m;
+    scanf("%d%d", &n, &m);
+
+    op1[0] = 1.0, op2[0] = 0.0;
+    for (int i = 1; i <= n; ++i) {
+        int type;
+        double x;
+        scanf("%d%lf", &type, &x);
+
+        if (type == 1) {
+            op1[i] = op1[i - 1] * x;
+            op2[i] = op2[i - 1];
+        } else {
+            op1[i] = op1[i - 1];
+            op2[i] = op2[i - 1] + x;
+        }
+    }
+    while (m--) {
+        int i, j;
+        double x, y;
+        scanf("%d%d%lf%lf", &i, &j, &x, &y);
+        double k = op1[j] / op1[i - 1];
+
+        x = k * x, y = k * y;
+        double t = op2[j] - op2[i - 1]; // 这里的t采用的弧度制，不用转化
+
+        // 注意这里计算ty的时候要用之前的x不能用tx,所以要再声明变量
+        double tx = x * cos(t) - y * sin(t);
+        double ty = x * sin(t) + y * cos(t);
+
+        printf("%.3lf %.3lf\n", tx, ty);
+    }
+
+    return 0;
+}

CCF-CSP/CSP202104A. 灰度直方图.cpp

@@ -0,0 +1,35 @@
+/*
+ * @Author: tkzzzzzz6
+ * @Date: 2026-05-26 21:45:57
+ * @LastEditors: tkzzzzzz6
+ * @LastEditTime: 2026-05-26 21:53:28
+ */
+#include <iostream>
+#include <vector>
+
+using namespace std;
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int n, m, l;
+    cin >> n >> m >> l;
+    vector<int> h(l, 0);
+
+    for (int i = 0; i < n; ++i)
+        for (int j = 0; j < m; ++j) {
+            int x;
+            cin >> x;
+            h[x]++;
+        }
+
+    for (int i = 0; i < l; ++i) {
+        if (i) cout << ' ';
+        cout << h[i];
+    }
+
+    cout << '\n';
+
+    return 0;
+}

CCF-CSP/[CSPro 31] 坐标变换（其一）.cpp

@@ -0,0 +1,28 @@
+/*
+ * @Author: tkzzzzzz6
+ * @Date: 2026-05-26 22:00:56
+ * @LastEditors: tkzzzzzz6
+ * @LastEditTime: 2026-05-26 22:05:24
+ */
+#include <iostream>
+
+using namespace std;
+
+int main() {
+    int n, m;
+    cin >> n >> m;
+
+    int dx = 0, dy = 0;
+    while (n--) {
+        int x, y;
+        cin >> x >> y;
+        dx += x, dy += y;
+    }
+
+    while (m--) {
+        int x, y;
+        cin >> x >> y;
+        cout << x + dx << ' ' << y + dy << '\n';
+    }
+    return 0;
+}

leetcode/problems/3120.count-the-number-of-special-characters-i-2.py

@@ -0,0 +1,44 @@
+'''
+Author: tkzzzzzz6
+Date: 2026-05-26 21:21:16
+LastEditors: tkzzzzzz6
+LastEditTime: 2026-05-26 21:37:07
+'''
+#
+# @lc app=leetcode.cn id=3120 lang=python3
+# @lcpr version=30204
+#
+# [3120] 统计特殊字母的数量 I
+#
+
+
+# @lcpr-template-start
+
+# @lcpr-template-end
+# @lc code=start
+class Solution:
+    def numberOfSpecialChars(self, word: str) -> int:
+        mask = [0]*2
+        for c in map(ord,word):
+            mask[c >>  5 & 1] |= 1<<(c & 31)
+
+        return bin(mask[0] & mask[1]).count("1")
+# @lc code=end
+    
+
+
+
+#
+# @lcpr case=start
+# "aaAbcBC"\n
+# @lcpr case=end
+
+# @lcpr case=start
+# "abc"\n
+# @lcpr case=end
+
+# @lcpr case=start
+# "abBCab"\n
+# @lcpr case=end
+
+#

leetcode/problems/3120.count-the-number-of-special-characters-i.md

@@ -0,0 +1,60 @@
+<!--
+ * @Author: tkzzzzzz6
+ * @Date: 2026-05-26 21:26:02
+ * @LastEditors: tkzzzzzz6
+ * @LastEditTime: 2026-05-26 21:44:25
+-->
+# 3120. 统计特殊字母的数量 1
+
+## 题目描述
+
+给一个字符串 `word`。如果 `word` 中同时存在某个字母的小写形式和大写形式，则称这个字母为 **特殊字母** 。
+
+返回 `word` 中 **特殊字母** 的数量。
+
+## 思路1:求集合的交集
+构建两个set,分别表示出现的大写和小写的字母
+
+set 去除重复的元素再求两个集合的交集即可
+
+- 时间复杂度：O(n)
+- 空间复杂度：O(n)
+
+### 代码
+
+```py
+class Solution:
+    def numberOfSpecialChars(self, word: str) -> int:
+        upper  = set()
+        lower = set()
+
+        for c in word:
+            if c == c.lower():
+                lower.add(c)
+            else:
+                upper.add(c.lower())
+
+        return  len(upper & lower)
+```
+
+## 思路2：位运算
+
+注意到大小写字母的ASCII码差值是32,转化为2进制后判断第5位是否是1即可
+
+- A-Z:65-90
+- a-z:97-122
+
+两者之间的差别是32
+
+可以采用位运算压缩和存储信息，并使用与位运算求解交集
+
+### 代码
+```py
+class Solution:
+    def numberOfSpecialChars(self, word: str) -> int:
+        mask = [0]*2
+        for c in map(ord,word):
+            mask[c >>  5 & 1] |= 1<<(c & 31)
+
+        return bin(mask[0] & mask[1]).count("1")
+```

leetcode/problems/3120.count-the-number-of-special-characters-i.py

@@ -1,3 +1,9 @@
+'''
+Author: tkzzzzzz6
+Date: 2026-05-26 21:21:16
+LastEditors: tkzzzzzz6
+LastEditTime: 2026-05-26 21:25:35
+'''
 #
 # @lc app=leetcode.cn id=3120 lang=python3
 # @lcpr version=30204
@@ -12,8 +18,18 @@
 # @lc code=start
 class Solution:
     def numberOfSpecialChars(self, word: str) -> int:
-        upper = 
+        upper  = set()
+        lower = set()
+
+        for c in word:
+            if c == c.lower():
+                lower.add(c)
+            else:
+                upper.add(c.lower())
+
+        return  len(upper & lower)
 # @lc code=end
+