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update(docs,typo): 添加问题“3025.人员站位的方案数I”的代码和题解+文章Mac发烫 (#1109)
3025: AC.cpp+py+go (#1108) cpp - AC,28.00%,90.00% cpp(rewrite) - AC,26.00%,50.00% py - AC,13.64%,18.18% go - AC,57.14%,42.86% rust和java详见下一个分支吧 git stash push -m "hello from #1108" Codes/lib.rs Codes/*.java Codes/*.rs 保存工作目录和索引状态 On 3025: hello from #1108 Signed-off-by: LetMeFly666 <[email protected]>
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‎.commitmsg‎

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3025: AC.cpp+py+go (#1108)
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cpp - AC,28.00%,90.00%
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cpp(rewrite) - AC,26.00%,50.00%
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py - AC,13.64%,18.18%
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go - AC,57.14%,42.86%
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rust和java详见下一个分支吧
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git stash push -m "hello from #1108" Codes/lib.rs Codes/*.java Codes/*.rs
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保存工作目录和索引状态 On 3025: hello from #1108

‎Codes/2438-range-product-queries-of-powers.py‎

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -15,7 +15,7 @@ def productQueries(self, n: int, queries: List[List[int]]) -> List[int]:
1515
pows.append(1 << th)
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th += 1
1717
n >>= 1
18-
perfix = [1] * (len(pows) + 1)
18+
prefix = [1] * (len(pows) + 1)
1919
for i in range(1, len(pows) + 1):
20-
perfix[i] = perfix[i - 1] * pows[i - 1]
21-
return [perfix[q[1] + 1] // perfix[q[0]] % 1000000007 for q in queries]
20+
prefix[i] = prefix[i - 1] * pows[i - 1]
21+
return [prefix[q[1] + 1] // prefix[q[0]] % 1000000007 for q in queries]

‎Codes/3025-find-the-number-of-ways-to-place-people-i.cpp‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-09-02 13:08:07
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-09-02 13:45:45
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*/
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#if defined(_WIN32) || defined(__APPLE__)
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#include "_[1,2]toVector.h"
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#endif
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class Solution {
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private:
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inline bool check(vector<vector<int>>& points, int i, int j) { // 易错点3：单独的(i, j)也要check
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if (points[i][0] > points[j][0] || points[i][1] < points[j][1]) { // 易错点1：注意这里纵坐标大于等于才合法
15+
return false;
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}
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return true;
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}
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inline bool check(vector<vector<int>>& points, int i, int j, int k) {
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if (points[i][0] <= points[k][0] && points[k][0] <= points[j][0] && points[i][1] >= points[k][1] && points[k][1] >= points[j][1]) {
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return false;
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}
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return true;
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}
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public:
26+
int numberOfPairs(vector<vector<int>>& points) {
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int ans = 0;
28+
for (int i = 0; i < points.size(); i++) {
29+
for (int j = 0; j < points.size(); j++) {
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if (i == j) {
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continue;
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}
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if (!check(points, i, j)) {
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continue;
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}
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bool can = true; // 易错点2：有一个k导致不符就不符
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for (int k = 0; k < points.size(); k++) {
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if (k == i || k == j) {
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continue;
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}
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if (!check(points, i, j, k)) {
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can = false;
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break;
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}
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}
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ans += can;
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}
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}
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return ans;
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}
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};
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#if defined(_WIN32) || defined(__APPLE__)
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/*
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[[1,1],[2,2],[3,3]]
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0
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*/
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/*
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[[0,0],[0,3]]
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1
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*/
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/*
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[[6,2],[4,4],[2,6]]
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(2,1) (1, 0)
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2
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*/
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/*
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[[0,0],[0,3]]
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1
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*/
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int main() {
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string s;
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while (cin >> s) {
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vector<vector<int>> v = stringToVectorVector(s);
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Solution sol;
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cout << sol.numberOfPairs(v) << endl;
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}
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return 0;
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}
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#endif

‎Codes/3025-find-the-number-of-ways-to-place-people-i.go‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-09-02 13:08:07
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-09-02 18:53:07
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*/
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package main
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func check2_3025(points [][]int, i, j int) bool {
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return points[i][0] <= points[j][0] && points[i][1] >= points[j][1]
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}
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func check3_3025(points [][]int, i, j, k int) bool {
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return !(points[i][0] <= points[k][0] && points[k][0] <= points[j][0] &&
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points[i][1] >= points[k][1] && points[k][1] >= points[j][1])
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}
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func numberOfPairs(points [][]int) (ans int) {
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for i := range points {
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for j := range points {
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if i == j {
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continue
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}
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if !check2_3025(points, i, j) {
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continue
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}
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can := true
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for k := range points {
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if k == i || k == j {
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continue
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}
32+
if !check3_3025(points, i, j, k) {
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can = false
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break
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}
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}
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if can {
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ans++
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}
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}
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}
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return
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}

‎Codes/3025-find-the-number-of-ways-to-place-people-i.py‎

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'''
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Author: LetMeFly
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Date: 2025-09-02 13:08:07
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LastEditors: LetMeFly.xyz
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LastEditTime: 2025-09-02 18:47:49
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'''
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from typing import List
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class Solution:
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def check2(self, i: int, j: int) -> bool:
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return self.points[i][0] <= self.points[j][0] and self.points[i][1] >= self.points[j][1]
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def check3(self, i: int, j: int, k: int) -> bool:
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return not (self.points[i][0] <= self.points[k][0] <= self.points[j][0] and self.points[i][1] >= self.points[k][1] >= self.points[j][1])
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def numberOfPairs(self, points: List[List[int]]) -> int:
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ans = 0
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n = len(points)
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self.points = points
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for i in range(n):
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for j in range(n):
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if i == j:
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continue
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if not self.check2(i, j):
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continue
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can = True
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for k in range(n):
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if k == i or k == j:
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continue
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if not self.check3(i, j, k):
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can = False
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break
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ans += can
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return ans

‎Codes/3025-find-the-number-of-ways-to-place-people-i_again.cpp‎

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/*
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* @Author: LetMeFly
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* @Date: 2025-09-02 13:08:07
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* @LastEditors: LetMeFly.xyz
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* @LastEditTime: 2025-09-02 13:51:14
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* @Description: rewrite from 0
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*/
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#if defined(_WIN32) || defined(__APPLE__)
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#include "_[1,2]toVector.h"
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#endif
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class Solution {
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private:
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inline bool check(vector<vector<int>>& points, int i, int j) {
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return i != j && points[i][0] <= points[j][0] && points[i][1] >= points[j][1];
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}
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inline bool check(vector<vector<int>>& points, int i, int j, int k) {
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return !(points[i][0] <= points[k][0] && points[k][0] <= points[j][0]
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&& points[i][1] >= points[k][1] && points[k][1] >= points[j][1]);
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}
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public:
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int numberOfPairs(vector<vector<int>>& points) {
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int ans = 0;
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for (int i = 0; i < points.size(); i++) {
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for (int j = 0; j < points.size(); j++) {
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if (!check(points, i, j)) {
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continue;
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}
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bool can = true;
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for (int k = 0; k < points.size(); k++) {
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if (k == i || k == j) {
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continue;
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}
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if (!check(points, i, j, k)) {
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can = false;
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break;
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}
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}
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ans += can;
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}
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}
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return ans;
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}
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};
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‎README.md‎

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|3011.判断一个数组是否可以变为有序|中等|<a href="https://leetcode.cn/problems/find-if-array-can-be-sorted/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/07/13/LeetCode%203011.%E5%88%A4%E6%96%AD%E4%B8%80%E4%B8%AA%E6%95%B0%E7%BB%84%E6%98%AF%E5%90%A6%E5%8F%AF%E4%BB%A5%E5%8F%98%E4%B8%BA%E6%9C%89%E5%BA%8F/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/140391465" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-if-array-can-be-sorted/solutions/2841665/letmefly-3011pan-duan-yi-ge-shu-zu-shi-f-i9ck/" target="_blank">LeetCode题解</a>|
922922
|3019.按键变更的次数|简单|<a href="https://leetcode.cn/problems/number-of-changing-keys/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/01/07/LeetCode%203019.%E6%8C%89%E9%94%AE%E5%8F%98%E6%9B%B4%E7%9A%84%E6%AC%A1%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144983704" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/number-of-changing-keys/solutions/3040949/letmefly-3019an-jian-bian-geng-de-ci-shu-pgzx/" target="_blank">LeetCode题解</a>|
923923
|3024.三角形类型|简单|<a href="https://leetcode.cn/problems/type-of-triangle/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/05/19/LeetCode%203024.%E4%B8%89%E8%A7%92%E5%BD%A2%E7%B1%BB%E5%9E%8B/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/148061722" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/type-of-triangle/solutions/3680826/letmefly-3024san-jiao-xing-lei-xing-shou-q46h/" target="_blank">LeetCode题解</a>|
924+
|3025.人员站位的方案数I|中等|<a href="https://leetcode.cn/problems/find-the-number-of-ways-to-place-people-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/09/02/LeetCode%203025.%E4%BA%BA%E5%91%98%E7%AB%99%E4%BD%8D%E7%9A%84%E6%96%B9%E6%A1%88%E6%95%B0I/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/151119511" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-the-number-of-ways-to-place-people-i/solutions/3770378/letmefly-3025ren-yuan-zhan-wei-de-fang-a-02tk/" target="_blank">LeetCode题解</a>|
924925
|3033.修改矩阵|简单|<a href="https://leetcode.cn/problems/modify-the-matrix/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/07/05/LeetCode%203033.%E4%BF%AE%E6%94%B9%E7%9F%A9%E9%98%B5/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/140219034" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/modify-the-matrix/solutions/2832430/letmefly-3033xiu-gai-ju-zhen-yuan-di-xiu-e0cy/" target="_blank">LeetCode题解</a>|
925926
|3038.相同分数的最大操作数目I|简单|<a href="https://leetcode.cn/problems/maximum-number-of-operations-with-the-same-score-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/06/07/LeetCode%203038.%E7%9B%B8%E5%90%8C%E5%88%86%E6%95%B0%E7%9A%84%E6%9C%80%E5%A4%A7%E6%93%8D%E4%BD%9C%E6%95%B0%E7%9B%AEI/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/139535702" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-number-of-operations-with-the-same-score-i/solutions/2804245/letmefly-3038xiang-tong-fen-shu-de-zui-d-dkj7/" target="_blank">LeetCode题解</a>|
926927
|3046.分割数组|简单|<a href="https://leetcode.cn/problems/split-the-array/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/12/28/LeetCode%203046.%E5%88%86%E5%89%B2%E6%95%B0%E7%BB%84/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144789679" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/split-the-array/solutions/3032725/letmefly-3046fen-ge-shu-zu-ji-shu-by-tis-33vv/" target="_blank">LeetCode题解</a>|

‎Solutions/LeetCode 1616.分割两个字符串得到回文串.md‎

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7474
$newStr$由三部分组成：$abkfkba = ab + kfk + ba$
7575

76-
其中$s_1 = ab \in a_{perfix}$，$s_3 = ba \in b_{suffix}$，$ab$和$ba$互为回文串，$s_2 = kfk$自身为回文串。
76+
其中$s_1 = ab \in a_{prefix}$，$s_3 = ba \in b_{suffix}$，$ab$和$ba$互为回文串，$s_2 = kfk$自身为回文串。
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7878
**那么思路来了：**
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8080
一个指针指向$a$串的首部，另一个指针指向$b$串的尾部，当两个指针所指字符相等时，a指针后移b指针前移，直到两指针相遇或两指针所指不同为止。
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82-
+ 如果两指针相遇，则说明$a_{perfix}$和$b_{suffix}$已经互为回文，$s_2$为空即可，直接返回$true$
82+
+ 如果两指针相遇，则说明$a_{prefix}$和$b_{suffix}$已经互为回文，$s_2$为空即可，直接返回$true$
8383
+ 如果两指针所指不同，则a指针前面的部分视为$s_1$，b指针后面的部分视为$s_3$（可以保证$s_1$和$s_3$互为回文），字符串a**或**字符串b 从a指针到b指针的部分 视为$s_2$，只需要判断$s_2$自身是否为回文串即可。若是则返回true，不是则返回false
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85-
上面判断了$a_{perfix} + b_{suffix}$的情况，$b_{perfix} + a_{suffix}$则同理
85+
上面判断了$a_{prefix} + b_{suffix}$的情况，$b_{prefix} + a_{suffix}$则同理
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8787
+ 时间复杂度$O(len(a))$
8888
+ 空间复杂度$O(1)$

‎Solutions/LeetCode 2438.二的幂数组中查询范围内的乘积.md‎

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两种方法：
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1. 直接暴力从query[0]累乘到query[1]并随时取模
64-
2. 前缀和的思想，使用数组perfix[j]代表$\prod_{0}^{j-1}powers[i]$
64+
2. 前缀和的思想，使用数组prefix[j]代表$\prod_{0}^{j-1}powers[i]$
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6666
直接暴力的方法并不会很耗时，因为powers数组中最多30个元素；
6767

@@ -138,10 +138,10 @@ class Solution:
138138
pows.append(1 << th)
139139
th += 1
140140
n >>= 1
141-
perfix = [1] * (len(pows) + 1)
141+
prefix = [1] * (len(pows) + 1)
142142
for i in range(1, len(pows) + 1):
143-
perfix[i] = perfix[i - 1] * pows[i - 1]
144-
return [perfix[q[1] + 1] // perfix[q[0]] % 1000000007 for q in queries]
143+
prefix[i] = prefix[i - 1] * pows[i - 1]
144+
return [prefix[q[1] + 1] // prefix[q[0]] % 1000000007 for q in queries]
145145
```
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> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/150227536)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2025/08/11/LeetCode%202438.%E4%BA%8C%E7%9A%84%E5%B9%82%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9F%A5%E8%AF%A2%E8%8C%83%E5%9B%B4%E5%86%85%E7%9A%84%E4%B9%98%E7%A7%AF/)哦~

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