第一题
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.gitignore
vendored
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2
.gitignore
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.idea/
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build_C/
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9
CMakeLists.txt
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CMakeLists.txt
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cmake_minimum_required(VERSION 3.28)
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project(lico)
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include_directories(Solving)
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aux_source_directory(Solving SOL_LIST)
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set(CMAKE_CXX_STANDARD 11)
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add_executable(lico main.cpp ${SOL_LIST})
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7
Solving/Questions.h
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Solving/Questions.h
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#ifndef LICO_QUESTIONS_H
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#define LICO_QUESTIONS_H
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//1、字符串分割
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void StringSegmentation();
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#endif //LICO_QUESTIONS_H
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57
Solving/StringSegmentation.cpp
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Solving/StringSegmentation.cpp
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#include "Questions.h"
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/**
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给定一个非空字符串S,其被N个‘-’分隔成N+1的子串,给定正整数K,要求除第一个子串外,
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其余的子串每K个字符组成新的子串,并用‘-’分隔。对于新组成的每一个子串,如果它含有的
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小写字母比大写字母多,则将这个子串的所有大写字母转换为小写字母;反之,如果它含有的
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大写字母比小写字母多,则将这个子串的所有小写字母转换为大写字母;大小写字母的数量相等时,不做转换。
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输入描述:
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输入为两行,第一行为参数K,第二行为字符串S。
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输出描述:
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输出转换后的字符串。
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示例1
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输入
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3
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12abc-abCABc-4aB@
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输出
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12abc-abc-ABC-4aB-@
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*/
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#include <iostream>
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void StringSegmentation() {
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std::string inputStr;
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int inputNmb;
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std::cin >> inputNmb;
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std::cin >> inputStr;
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std::string output;
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int index = inputStr.find('-');
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if (index == -1) {
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return;
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}
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output = std::string(inputStr.begin(), inputStr.begin() + index + 1);
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inputStr = inputStr.substr(index + 1);
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std::string tmpStr{};
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index = inputStr.find('-');
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while (index != -1) {
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tmpStr += inputStr.substr(0, index);
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inputStr = inputStr.substr(index + 1);
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index = inputStr.find('-');
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}
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tmpStr += inputStr;
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for (auto i: tmpStr) {
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int low{}, up{}, cnt{};
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std::string lowStr = "-";
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std::string upStr = "-";
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if (i >= 'a' && i <= 'z') {
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++low;
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lowStr.push_back(i);
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upStr.push_back(i + 26);
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}
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else if (i >= 'A' && i <= 'z')
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}
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std::cout << output;
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}
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