UVaOJ 401 – Palindromes

问题描述

解法一：模拟

模拟读入，每次读一行字符串 s ，然后判断是否是回文串或镜像串：

回文串：对于每个 s[i] 都必须满足 s[i]==s[len-i-1], 这个验证起来比较简单；
镜像串：对于每个 s[i] 都必须满足 FindMirror(s[i])==s[len-i-1], 需要知道如何设计 FindMirror 函数

我们可以借助字符串常量 MIRR_ALPHAS 和 MIRR_NUMBER 来建立字符（字母和数字）到其镜像字符的映射，实现上用不同字符编码的相对偏移量当作对应索引值。同时这里在处理输出字符串时用到了一些技巧，由于回文串和镜像串都是简单的 0/1 关系，所以可以用二进制来映射所有的 NUM_CASE = 4 种情况。

容易忽视的地方：此题的输出需要在输入字符串后面接内容，因此在从标准流读入的时候记得关注换行符的处理。

#include <bits/stdc++.h>

using namespace std;

// For comparison:        "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
char const* MIRR_ALPHAS = "A   3  HIL JM O   2TUVWXY5";
char const* MIRR_NUMBER = "1SE Z  8 ";
//                        "123456789"

const int NUM_CASE = 4;
const array<string, NUM_CASE> SUFFIX = {
    " -- is not a palindrome.", " -- is a regular palindrome.",
    " -- is a mirrored string.", " -- is a mirrored palindrome."};

char FindMirror(const char& ch) {
  if (isalpha(ch))
    return MIRR_ALPHAS[ch - 'A'];
  else  // isdigit
    return MIRR_NUMBER[ch - '1'];
}

int main() {
  string s;
  while (getline(cin, s)) {
    int len = s.size();
    bool is_palindrome = 1, is_mirrored = 1;
    for (int i = 0; (is_palindrome || is_mirrored) && i < (len + 1) / 2; ++i) {
      char mirr_char = FindMirror(s[i]);
      if (mirr_char != s[len - i - 1]) is_mirrored = 0;
      if (s[i] != s[len - i - 1]) is_palindrome = 0;
    }
    cout << s + SUFFIX[is_palindrome + is_mirrored * 2] << "\n\n";
  }
}

#include <bits/stdc++.h>

using namespace std;

// For comparison: "ABCDEFGHIJKLMNOPQRSTUVWXYZ"

char const* MIRR_ALPHAS = "A 3 HIL JM O 2TUVWXY5";

char const* MIRR_NUMBER = "1SE Z 8 ";

// "123456789"

const int NUM_CASE = 4;

const array<string, NUM_CASE> SUFFIX = {

" -- is not a palindrome.", " -- is a regular palindrome.",

" -- is a mirrored string.", " -- is a mirrored palindrome."};

char FindMirror(const char& ch) {

if (isalpha(ch))

return MIRR_ALPHAS[ch - 'A'];

else // isdigit

return MIRR_NUMBER[ch - '1'];

}

int main() {

string s;

while (getline(cin, s)) {

int len = s.size();

bool is_palindrome = 1, is_mirrored = 1;

for (int i = 0; (is_palindrome || is_mirrored) && i < (len + 1) / 2; ++i) {

char mirr_char = FindMirror(s[i]);

if (mirr_char != s[len - i - 1]) is_mirrored = 0;

if (s[i] != s[len - i - 1]) is_palindrome = 0;

}

cout << s + SUFFIX[is_palindrome + is_mirrored * 2] << "\n\n";

}

MIRR_ALPHAS = "A   3  HIL JM O   2TUVWXY5"
MIRR_NUMBER = "1SE Z  8 "

SUFFIX = [" -- is not a palindrome.",
          " -- is a regular palindrome.",
          " -- is a mirrored string.",
          " -- is a mirrored palindrome."]


def find_mirr_char(ch):
    return MIRR_ALPHAS[ord(ch) - ord("A")] if ch.isalpha() else MIRR_NUMBER[ord(ch) - ord("1")]

while True:
    try:
        line = input()
        is_palindrome = 1
        is_mirrored = 1
        for i, _ in enumerate(line):
            mirr_char = find_mirr_char(line[i])
            if mirr_char != line[len(line) - i - 1]:
                is_mirrored = 0
            if line[i] != line[len(line) - i - 1]:
                is_palindrome = 0
        print(line + SUFFIX[is_palindrome + is_mirrored * 2], end="\n\n")
    except EOFError:
        break

MIRR_ALPHAS = "A 3 HIL JM O 2TUVWXY5"

MIRR_NUMBER = "1SE Z 8 "

SUFFIX = [" -- is not a palindrome.",

" -- is a regular palindrome.",

" -- is a mirrored string.",

" -- is a mirrored palindrome."]

def find_mirr_char(ch):

return MIRR_ALPHAS[ord(ch) - ord("A")] if ch.isalpha() else MIRR_NUMBER[ord(ch) - ord("1")]

while True:

try:

line = input()

is_palindrome = 1

is_mirrored = 1

for i, _ in enumerate(line):

mirr_char = find_mirr_char(line[i])

if mirr_char != line[len(line) - i - 1]:

is_mirrored = 0

if line[i] != line[len(line) - i - 1]:

is_palindrome = 0

print(line + SUFFIX[is_palindrome + is_mirrored * 2], end="\n\n")

except EOFError:

break

AOAPC II UVaOJ 字符串

UVaOJ 401 - Palindromes

问题描述

解法一：模拟