我的PAT-BASIC代碼倉:https://github.com/617076674/PAT-BASIC
原題連結:
PAT-BASIC1044:https://pintia.cn/problem-sets/994805260223102976/problems/994805279328157696
PAT-ADVANCED1100:https://pintia.cn/problem-sets/994805342720868352/problems/994805367156883456
題目描述:
PAT-BASIC1044:
PAT-ADVANCED1100:
知識點:進制轉換
思路:用數組存儲火星文,轉成13進制即可
注意點:
如果末位是0,不需要輸出“tret”,除非該數字就是0。
0應該轉換成“tret”。13應該轉換成“tam”,而不是“tam tret”。
由于輸入數字在[0, 169)區間内,是以分析時間複雜度和空間複雜度意義不大。
C++代碼:
#include<iostream>
#include<string>
#include<sstream>
using namespace std;
int N;
string lowStrings[13] = {"tret", "jan", "feb", "mar", "apr", "may", "jun", "jly", "aug", "sep", "oct", "nov", "dec"};
string highStrings[12] = {"tam", "hel", "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok", "mer", "jou"};
int marToEarth(string s);
string earthToMar(int num);
int main() {
scanf("%d", &N);
getchar(); //讀取換行符
string tempNum;
int num;
for (int i = 0; i < N; i++) {
getline(cin, tempNum);
if (tempNum[0] >= '0' && tempNum[0] <= '9') {
stringstream transfer;
transfer << tempNum;
transfer >> num;
cout << earthToMar(num) << endl;
} else {
cout << marToEarth(tempNum) << endl;
}
}
}
int marToEarth(string s) {
if (s.length() <= 4) {
for (int i = 0; i < 13; i++) {
if (s.compare(lowStrings[i]) == 0) {
return i;
}
}
for (int i = 0; i < 12; i++) {
if (s.compare(highStrings[i]) == 0) {
return (i + 1) * 13;
}
}
} else {
int result = 0;
for (int i = 0; i < 12; i++) {
if (s.substr(0, 3).compare(highStrings[i]) == 0) {
result += (i + 1) * 13;
break;
}
}
for (int i = 0; i < 13; i++) {
if (s.substr(4, 3).compare(lowStrings[i]) == 0) {
result += i;
break;
}
}
return result;
}
}
string earthToMar(int num) {
if (num <= 12) {
return lowStrings[num];
} else {
string result = "";
result += highStrings[num / 13 - 1];
if (num % 13 != 0) {
result += " ";
result += lowStrings[num % 13];
}
return result;
}
}
C++解題報告: