UVaOJ 1339 - Ancient Cipher

问题描述

解法一：统计后排序

统计两个串中各个大写字母的出现次数，如果能够通过 substitution cipher 和 permutation cipher 加密解密，表明一定能找到 26 个字母之间的一一映射，且满足出现次数一致。因此这里对字母出现次数做一个排序，如果发现有出现次数无法匹配的情况，表明该字母找不到对方串中的字母进行映射，输出 NO.

#include <bits/stdc++.h>

using namespace std;

int main() {
  const int ALPHABET_NUM = 26;
  string s1, s2;
  while (cin >> s1 >> s2) {
    array<int, ALPHABET_NUM> cnt1 = {0}, cnt2 = {0};
    for (auto c : s1) cnt1[c - 'A']++;
    for (auto c : s2) cnt2[c - 'A']++;
    sort(cnt1.begin(), cnt1.end());
    sort(cnt2.begin(), cnt2.end());
    bool encrypt = true;
    for (int i = 0; i < ALPHABET_NUM; ++i)
      if (cnt1[i] != cnt2[i]) {
        encrypt = false;
        break;
      }
    cout << (encrypt ? "YES" : "NO") << endl;
  }
  return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main() {

const int ALPHABET_NUM = 26;

string s1, s2;

while (cin >> s1 >> s2) {

array<int, ALPHABET_NUM> cnt1 = {0}, cnt2 = {0};

for (auto c : s1) cnt1[c - 'A']++;

for (auto c : s2) cnt2[c - 'A']++;

sort(cnt1.begin(), cnt1.end());

sort(cnt2.begin(), cnt2.end());

bool encrypt = true;

for (int i = 0; i < ALPHABET_NUM; ++i)

if (cnt1[i] != cnt2[i]) {

encrypt = false;

break;

}

cout << (encrypt ? "YES" : "NO") << endl;

}

return 0;

}

ALPHABET_NUM = 26
while True:
    try:
        s1, s2 = input(), input()
        cnt1 = [0 for _ in range(ALPHABET_NUM)]
        cnt2 = [0 for _ in range(ALPHABET_NUM)]
        for c in s1:
            cnt1[ord(c) - ord('A')] += 1
        for c in s2:
            cnt2[ord(c) - ord('A')] += 1
        cnt1, cnt2 = sorted(cnt1), sorted(cnt2)
        print("YES" if cnt1 == cnt2 else "NO")
    except EOFError:
        break

ALPHABET_NUM = 26

while True:

try:

s1, s2 = input(), input()

cnt1 = [0 for _ in range(ALPHABET_NUM)]

cnt2 = [0 for _ in range(ALPHABET_NUM)]

for c in s1:

cnt1[ord(c) - ord('A')] += 1

for c in s2:

cnt2[ord(c) - ord('A')] += 1

cnt1, cnt2 = sorted(cnt1), sorted(cnt2)

print("YES" if cnt1 == cnt2 else "NO")

except EOFError:

break

AOAPC II UVaOJ 字符串