AtCode249 - B - Perfect String

题目地址

B - Perfect String

• https://atcoder.jp/contests/abc249/tasks/abc249_b

問題描述

Problem Statement

Let us call a string consisting of uppercase and lowercase English alphabets a wonderful string if all of the following conditions are satisfied:

• The string contains an uppercase English alphabet.
• The string contains a lowercase English alphabet.
• All characters in the string are pairwise distinct.

For example, AtCoder and Aa are wonderful strings, while atcoder and Perfect are not.

Given a string SS, determine if SS is a wonderful string.

Constraints

• 1≤∣S∣≤100
• S is a string consisting of uppercase and lowercase English alphabets.

---

Input

Input is given from Standard Input in the following format:

S

Output

If SS is a wonderful string, print Yes; otherwise, print No.

---

Sample Input 1

AtCoder

Sample Output 1

Yes

AtCoder is a wonderful string because it contains an uppercase English alphabet, a lowercase English alphabet, and all characters in the string are pairwise distinct.

---

Sample Input 2

Aa

Sample Output 2

Yes

Note that A and a are different characters. This string is a wonderful string.

---

Sample Input 3

atcoder

Sample Output 3

No

It is not a wonderful string because it does not contain an uppercase English alphabet.

---

Sample Input 4

Perfect

Sample Output 4

No

It is not a wonderful string because the 22-nd and the 55-th characters are the same.

题意

• 完美字符串满足
  • 只包含英文
  • 至少包含一个大写和小写
  • 所有文字不重复

思路

• 循环遍历，定义计数器，使用c++自带的函数统计大写和小写字母数量， 均不为0则满足条件；
  • 不足：这块定义一个bool变量更优
• 循环时将每位字符存入set结构，因为set里的元素有互异性，所以最后只需要判断set的size是否和输入文字数一致即可。
  • 不足：空间复杂度上升，但处理简单，无脑就把程序写完

题解

小码匠

void coder_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    string s;
    cin >> s;
    set<char> a;
    int big = 0;
    int small = 0;
    for (int i = 0; i < s.size(); i++) {
        if (isupper(s[i])) {
            big++;
        } else if (islower(s[i])){
            small++;
        }
        a.insert(s[i]);
    }
    if (a.size() == s.size() && big > 0 && small > 0) {
        cout << "Yes";
    } else {
        cout << "No";
    }
}

参考题解

void second_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    string s;
    cin >> s;

    sort(s.begin(), s.end());
    for (int i = 0; i < s.length() - 1; i++) {
        if (s[i] == s[i + 1]) {
            cout << "No";
        }
    }

    bool a = false, b = false;
    for (auto &&c: s) {
        if (islower(c)) a = true;
        if (isupper(c)) b = true;
    }
    if (a && b) {
        cout << "Yes" << endl;
    }
}

官方题解

#include <bits/stdc++.h>
using namespace std;
int main() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    string s;
    cin >> s;
    bool big = false, small = false;
    for (int i = 0; i < s.size(); i++) {
        if (isupper(s[i])) big = true;
        else small = true;
    }
    bool diff = true;
    for (int i = 0; i < s.size(); i++) {
        for (int j = i + 1; j < s.size(); j++) {
            if (s[i] == s[j]) diff = false;
        }
    }
    if (big && small && diff) {
        cout << "Yes" << endl;
    } else {
        cout << "No" << endl;
    }
}

• 仔细分析官方题解，是有可优化空间的。
  • 第一：第8行、9行的判断中，如果bit和small都不为true，其实是没必要走后面循环处理；
  • 第二：第12行、13行用双指针来处理，更好。

待补知识点

• 依然是函数这块，需要补充。当时看到这个题解，把代码复制到IDE中

  • 一：编译未通过，应该是自己知识水平不够，大神提交代码是没问题，功力不够；

  • 二：模板那块是自己的软肋，还没学习。

  template <class T> T scan() {
      T ret;
      std::cin >> ret;
      return ret;
  }
  
  void best_solution() {
      // 提升cin、cout效率
      ios::sync_with_stdio(false);
      cin.tie(0);
      cout.tie(0);
  
      const std::string s = scan<std::string>();
  
      if (s.size() != set(s.begin(), s.end()).size()) {
          std::cout << "No" << endl;
          return;
      }
      bool upp = false;
      bool low = false;
      for (const char c: s) {
          if ('A' <= c && c <= 'Z') {
              upp = true;
          }
          if ('a' <= c && c <= 'z') {
              low = true;
          }
      }
      std::cout << (upp && low ? "Yes" : "No") << endl;
  }